School Mathematics Study Group
Geometry
Unit 1 3
Geometry Student's Text, Part I
Prepared under the supervision of the Panel on Sample Textbooks of the School ~ a t h e k t i c sStudy Group: Frank 3.Alien
Edwin C. Douglas Donald E. Richmond Charles E. Rickart Henry Swain Robert J. Walker
Lyons Township High School Tafi School Williams College Yale University New Trier Township High School Cornell University
New Haven and London,Yale University Press
Copyright @ 1960, 1961 by Yale University. Printed in the United States of America.
All rights reserved. This book may not be reproduced, in whole or in pan, in any form, without written permission from the publishers. Financial support for the School Mathematics Study Group has been provided by the National Science Foundation.
FOREWORD The Increasing c o n t r i b u t i o n of mathematics t o t h e culture of the modem world, as well as i t s Importance as a vital part of s c i e n t i f i c and humanistic education, has made it essential t h a t the mathematics I n o u r schools be both well selected and well taught.
With t h i s in mind, t h e various mathematical organizations i n the United States cooperated In t h e formation of t h e School Mathematics Study Group (SMSG) 3MSG includes college and unlver* s i t y mathematicians, teachers of mathematics a t a l l levels, experts in education, and representatives o f science and technology. The general o b j e c t i v e of SMSG is the improvement of the teaching of mathematics In the schools of t h i s country. The National Science Foundation has provided substantial funds f o r t h e s u p p o r t of this . endeavor.
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One of the prerequisites for the Improvement of the teaching of mathematics in our schools I s an improved curriculum--one which takes account of the Increasing use of mathematics in science and technology and in other areas of knowledge and at the same time one which r e f l e c t s recent advances I n mathematics itself. One of the first projects undertaken by SMSG was to e n l i s t a group of outstanding mathemat ic iana and mathematics teachers t o prepare a series of textbooks which would Illustrate such an improved
curriculum. The professional mathematicians in SMSG believe that t h e
mathematics presented l a t h i s t e x t i s valuable for all well-educated citizens i n our s o c i e t y to know and that it is important f o r the precollege student to learn in preparation f o r advanced work In t h e f i e l d . A t the same time, teachers in SMSG b e l i e v e that it is presented in such a .form that it c a n be readily grasped by students. In most instances t h e material w i l l have a familiar note, b u t the presentation and t h e p o i n t of view w i l l be d i f f e r e n t . Some material w i l l be e n t i r e l y new to the traditional curriculum. This I s a8 I t should be, for mathematics is a living and an ever-growing subject, and n o t a dead and frozen product of a n t i q u i t y . T h i s healthy f u s i o n of the o l d and the new should lead students to a better understanding of t h e basic concepts and structure of mathematics and provide a firmer foundation f o r understanding and use o f mathematics in a scientific s o c i e t y .
It is not intended t h a t this book be regarded as t h e o n l y definitive way o f presenting good mathematics t o students a t t h i s level. Instead, I t should be thought of as a sample of t h e kind of improved curriculwn that we need and as a source of suggestions for the authors of commercial textbooks. It I s sincerely hoped t h a t these t e x t s will lead the way toward Inspiring a more meaningful teaching of Mathematics, the Queen and Servant of the Sciences.
Below are H a t e d t h e names of all those who p a r t i c i p a t e d i n any of t h e writing sessions
at
whlch t h e f - o ~ ~ o w l nSMSG g t e x t s were prepared:
First Coarse i n Ugebra, Geometry,
IntemedLate Mathematics, Elementary Functions> and ~ n t ~ o d u c t i ot no Matrix Algebra.
H.W. Alexander, E a ~ l h a nCollege F.B. Allen, Lyons Township High School, La Grange, I l l i n o i s Alexander Beck, Oiney High School, Philadelphia, Pennsylvania 3 3 . Beckenbach, University o f C a l i f o r n i a a t LOB Angeles E .G. Begle, School Mathematics Study Group, Yale Vnlverslty Paul Berg, Stanford University Bull Berger, Monroe High School, S t . Paul, Minnesota Arthur Bemihart, University of Oklahoma R.H. Bing, University of Wisconsin A . L . Blalcers, University of Western Australia A . A . Blank, New York University S h i r l e y Boselly, Franklin High School, S e a t t l e , Washington K.E. -Brown, Department of Health, Eciacat i o n , and Welfare, Washington) D . C . 3 .M. Galloway, Carlet on College Hope Chipman, University High School, Ann Arbor, Michigan R.R. Christian, University o f B r i t i s h c o l umbia R.J. Clark, S t . Paul's School, Concord, New Hampshire P . H . Daus, University of C a l i f o r n i a a t Los Angeles R.B. Davis, Syracuse University Charles DePrima, California I n s t i t u t e of Technology Mary Dolclanl, Hunter College Edwin C . Douglas, The T a f t School, Watertown, Connecticut Floyd Downs, East High School, Denver, Colorado E .A, Dudley, North Haven High School, Worth Haven, Connecticut Lincoln Onrst, The Rice Institute Florence Elder, West Hempstead High School, West Hempstead, Hew York W.E. Ferguaon, Newton High School, Menrtonv i l l e , Massachusetts N.J. Pine, University of Pennsylvania Joyce 0 , Fontalne, North Haven High School, North Haven, Connecticut F.L. Friedman, Massachusetts I n s t i t u t e of Technology Esther Gassett, Claremore High School,
Claremore, Oklahoma
R.K. Getoor, University of Washington V.H. Haag, Franklin and Marshall College R .R Hartman, Rdina-Morningside Senior H i g h School, Edina, Minnesota M.H. Heine, University of I l l i n o i s Edwin Hewitt, University of Washington Martha Hlldebrandt, Proviso Township High School, Haywood, Ill lnoia
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Jurgensen, Culver Military Academy, Culver, Indiana Joseph Lehner, Michigan S t a t e University Marguerite Lehr, Bryn Maw College Kenneth Leisenrln&, University of Michigan Howard Levl., Columbia University Eunice Lewis, Laboratory High School, University o f Oklahoma M.A. Linton. WillSam Penn Charter School, Philadelphia, Pennsylvania A.E. Livingston, University of Washington L .H. Loomis, Harvard University R .V. Lynch, P h i l l l p s Exeter Academy, Exeter, New Hampshire W .K McMabb, Hoeleaday School, Dallas, Texas K.O. Michaels, North Haven High School, North Haven, Connecticut E.E. Moise, University of Michigan E . P . Northrop, University of Chicago O . J . Peterson, Kansas S t a t e Teachers College, h p o r i a , Kansas B . J . P e t t l s , University of North Carolina R S. P l e t e r s , P h l l l i p s Academy, Andover, Massachusetts H.O. Pollak, Bell Telephone Laboratories Walter Prenowltz, Brooklyn College O.B. Price, University of Kansas A.L. Putnam, University of Chicago P e r a l s 0. Redgrave, Norwich Free Academy, Norwich, Comec tiaut Kina Rees, Hunter College D.E. Richmond, Williams College C.E. Rickart, Yale University Harry Rwlerman, Hunter College High School, New York City J . T . Schwartz, New York University O.E. S t a n a i t i s , St. Olaf College Robert Starkey, Cubberley H i g h Schools, Palo Alto, California F h l l l i p Stucky, Roosevelt H i g h School, S e a t t l e , Washington Henry Swain, Mew Trier Township H i g h School, Vinnetka, I l l i n o i s Henry Syer, Kent School, Kent, Connecticut Q . 3 . Thomas, Massachusetts I n s t i t u t e of Technology A.M. Tucker, Princeton University H.E . Vaughan, University of ~ l l i n o i s John Wagner, University of Texas R . J . Walker. Cornell University A.D. Wallace, Tulane University E.L. Halters, Vllliam Penn Senior High School, York, Pennsylvania Warren White, North Hign School, Sheboygan, Misconaln D.V. Widder, Harvard University William Wooton, Pierce Junior College, Woodland H i l l s , California J.H. Zant, Oklahoma S t a t e University
R .C
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CONTENTS Chapter 1 COMMON SENSE AND ORGANIZED KNOWLEDGE 1 . Two Types ofProblems 1.2 An Organized Logical Development of Geometry
.
2.
........ ...........
.
. SETS. REALNUMBERSANDLINES .. ,. .. 2 . 1 . Sets . . . . . . . . . . . . . . . . . . . . 2 . . The Real Numbers . . . . . . . . . . . . . . ,
2-3 2-4
5 . 2
6 . 2 2 . 7
. .. . .
............. . . . .. .. .. .. .. .. .. ............. . . . . . .. .Befcweeness. ......
The Absolute Value Measurement of Distance A Choice of a Unit of Distance An Infinite Ruler The Ruler Placement Postulate Segments and Rays
. . . .. .. .. .a. ,. .. .. .. , ................. 3-3 . Convex S e t s . . . . . . . . . . . . . . . . 4 . ANGLES AND TRIANGLES . . . . . . . . . . . . . . . . 1 . 4 . The Basic Definitions . . . . . . . . . . . 42. . Remarks on Angles . . . . . . . . . . . . .
3 . LINES. PLANES AMD SEPARATION 3 . 1 5 . 2
43. 44.
..
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Lines and Planes i n S p a c e Theorems in t h e Form of Hypothesis and ~onclusion
...........
Measurement of Angles Perpendicularity. R i g h t Angles and Congruence of Angles ;
................ 5 . CONGRUENCES . . . . . . . . . . . . . . . . . . . . 5-1 . The Idea of a Congruence . . . . . . . . . . 5-2 . Congruences Between Triangles . . . . . . . 5.3 . The Basic Congruence P o s t u l a t e . . . . . . . 5.4 . Writing Your Own Proofs . . . . . . . . . . 55. . Overlapping Triangles . Using the F i g u r e i n Statements . . . . . . . . . . . . . . . . . 56. . The Isosceles Triangle Theorem . The Angle Bisector Theorem . . . . . . . . . . . . . . 5.7 . The Angle Side Angle Theorem . . . . . . . . 58 . . The Side Side Side Theorem . . . . . . . . . Review Exercises. Chapters 1 t o 5 . . . . . . . . . . . 6 . A CLOSER LOOK A T PROOF . . . . . . . . . . . . . . . 6 . 1 . How A Deductive System Works . . . . . . . . 6.2 . I n d i r e c t Proof . . . . . . . . . . . . . . . 6.5 . Theorems About Perpendiculars . . . . . . . . 6.4 . Introducing Auxiliary Sets Into Proofs . . . 65. . Betweenness and Separation . . . . . . . . .
1
8 15 15 21 27
^
30
39
53 53
60 62 71
71 77
79 85
97
97 109
115 117
123 127 132
137
155 159 159 160
167
176 182
Chapter
7
8
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9
10
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. . . . . . . . .. .. .. .. .. .. . . . .. .. .. .. .. .. ................
GEOMETRIC INEQUALITIES 7 . 1 MakineReasonable Con.lectures 7.2 ~ l g e b r aof 1nequaliti6s 7.3 The Basic I n e q u a l i t y Theorems 74 . Altitudes
.
.. .
. . . . . . .. .. .. .. .. 8.2 . The Basic Theorem . . . . . . . . . . . . 8-3 . Existence and Uniqueness Theorems . . . . PARALLEL LINES IN A PLANE . . . . . . . . . . . . 9 . 1 . Conditions Which Guarantee Parallelism . 9.2 . Corresponding Angles . . . . . . . . . . . 9.3 . The Parallel P o s t u l a t e . . . . . . . . . . 9.4 . Triangles . . . . . . . . . . . . . . . . 9.5 . Quadrilaterals in Plane . . . . . . . . . 9.6 . Rhombus. Rectangle and Square . . . . . . 9-7* Transversals To Many Parallel Lines . . . PERPENDICULAR LINES AND PLANES IN SPACE 8-1 The Basic D e f i n i t i o n
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. . . .. .. .. .. .. .. .. .. .. .. .. .. .. D i h e d r a l Angles. Perpendicular Planes . . Pro jectlons . . . . . . . . . . . . . . .
PARALLELS IN SPACE
.. .
10.1 10.2 10.3
.
Parallel Planes
..........
Appendix I A Convenient Shorthand Appendix I1 P o s t u l a t e s of Addition and Multiplication Appendix I11 Rational and I r r a t i o n a l Numbers Appendix IV Square Roots Appendix V How to Draw Figures in 3-Space Appendix VI Proofs of Theorems on P e r p e n d i c u l a r i t y THE MEANING AND USE OF SYMBOLS
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.
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..... ...............
...... . . ............. LIST OF POSTULATES . . . . . . . . . . . . . . . . . . . .
LIST OF THEOREMS AND COROLLARIES INDEX OF DEFINITIONS
189 189 191 19? 21
219 219 222 235 241 241 251 252
258 265 268 275
291 291 299
306 A-1
A-5 A-9
A-15 A-19 A-25 a
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e 1 y
PREFACE
This book is designed for the one-year i n t r o d u c t o r y course in geometry which is usually taught in the tenth grade. S t u d e n t s I n t h i s grade normally have learned a fair amount of Informal geometry, including the calculation of areas and volumes for various elementary figures , t h e Pythagorean r e l a tion, and the use of s i m i l a r r i g h t triangles t o c a l c u l a t e unknown h e i g h t s and dist a n c e s . Students who have not learned this m a t e r i a l may have to be given some extra a t t e n t i o n , but t h e book should s t i l l be teachable, a t a s u i t a b l y a d j u s t e d pace. I n algebra, no special p r e p a r a t i o n is required beyond the knowledge and skills normally acquired In the ninth grade. The book is devoted mainly to plane geometry, with a few chapters on solid geometry, and a s h o r t introduction to analytic geometry a t the end. It seems natural, in a preface, to give an account of the novel f e a t u r e s i n t h e treatment. W e a r e aware, of course, of a danger In 80 doing. A long string of novelties, offered for the reader's s p e c i a l a t t e n t i o n , may very well convey t h e impression that the authors have been engaged In an unhealthy pursuit of Innovation for i t s own sake. This is by no means the way in which we have conceived our task. We began and ended our work with t h e c o n v i c t i o n t h a t t h e traditional c o n t e n t of Euclidean geometry amply deserves t h e prominent place which It now holds In high-school study; and we have made changes only when t h e need for them appeared to be compelling. The basic scheme i n the postulates I s that of G. D. Birkhoff, In t h i s scheme, it i s assumed that the r e a l numbers are known, and they a r e used freely f o r measuring both d i s t a n c e s and angles. This has two main advantages.
In the first place, the real numbers give u s a sort of head It has been correctly pointed out t h a t Euclid's postulates a r e not l o g i c a l l y s u f f i c i e n t f o r geometry, and t h a t t h e treatments b a s e d on them do not m e e t modern standards of rigor. They were Improved a n d s h a r p e n e d by Hllbert. B u t the foundations of geometry, in the sense of Hilbert, a r e not a part of elementary mathematics, and do not belong in the tenth-grade curriculum. If we assume the real numbers, a s in the Birkhoff treatment, then the handling of o u r p o s t u l a t e s becomes a much e a s i e r t a s k , and we need n o t f a c e a c r u e l choice between mathematical accuracy and Intelligibility.
start.
In the second place, It seems a good idea i n I t s e l f t o connect up geometry with algebra a t every reasonable opportunity, so that knowledge in one of these fields will make its n a t u r a l cont r i b u t i o n to the understanding of both. Some of t h e topics u s u a l l y studied i n geometry are e s s e n t i a l l y algebraic. This i s true, for example, of the p r o p o r t i o n a l i t y r e l a t i o n s f o r similar triangles. In this book, such topics are treated algebraically, so a s to bring out the connections with the work of the ninth and eleventh grades.
We hope t h a t t h e statements of d e f i n i t i o n s and theorems are to make them so. Just as a lawyer needs t o learn to draw u p contracts that say what they a r e supposed to say, so a mathematics student needs to l e a r n t o write mathematic a l statements t h a t can b e taken literally. B u t we are not under the illusion t h a t this sort of e x a c t i t u d e I s a substitute for Intuitlve insight. We have, therefore, based t h e design of both t h e t e x t and the problems on our b e l i e f that Intuition a n d l o g i c s h o u l d move forward hand in hand, e x a c t ; we have t r i e d hard
Chapter 1
COMMON SENSE ANO ORGANIZED KNOWLEDGE 1-1.
Two Types -
of Problems.
Consider the following problems: 1. A line segment 14 inches long is broken I n t o two segments. If one of the two smaller segments is 6 inches long, how long is the other one? 2, In a c e r t a i n " r e c t a n g l e , t h e sum of the l e n g t h and the width is 14 (measured in inches). A second rectangle is three times as long as the first, and twice as wide. The perimeter of the second rectangle I s 72. What are the dimensions of the first rectangle? The answer to Problem 1, of course, is 8 inches, because 6 + 8 = 1 4 . We could solve this problem algebraically, if we wanted t o , by s e t t i n g up the equation 6 + x = 14, and solving to get x = 8. But this seems a little silly, because it is so unnecessary. If all algebraic equations were as superfluous as this one, then no serious-minded person would pay any attention to them; in fact, they would probably never have been invented. Problem 2, however, is quite another matter. If the length and width of the first rectangle are x and y, t h e n t h e length and width of the second rectangle a r e 3x and 2y. Therefore,
because t h e sum of the length and width is h a l f the perimeter. We already know t h a t x + y = 1 4 . Thus we have a sys tern of t w o l i n e a r equations In two unknowns; x + y = 1 4
3x
+3
= 36.
To solve, ive multiply each term In the first equation by 2, getting 2x + 2y = 28, and then we subtract this last equation, tern by term, from t h e second. This gives x = 8. Since x + y = 14, we have y = 6 , which completes the solution of our problem. It I s easy to check that a length of 8 and a width of 6 satisfy t h e c o n d i t i o n s o f t h e problem. In a way, .these t w o problems may seem similar. But in a very important sense, they are different. The first is what you might c a l l a common-sense problem. It is very easy to guess what the answer ought to be, and it is also very easy to check that the natural guess I s actually t h e right answer. The second problem is e n t i r e l y another matter. To solve the second problem, we need to know something about mathematical methods. There are cases of this kind in geometry. Consider the following statements: 1, If a triangle has s i d e s of length 3, 4 and 5, then i t is a r i g h t t r i a n g l e , with a right angle opposite the longest side. 2. L e t a triangle be given, with s i d e s a, b and c. If a 2 + b 2 = c2 , then the triangle is a right t r i a n g l e , with a right angle opposite t h e longest side. The first of these facts was known to the ancient Egyptians. They checked it by experiment. You c a n check it yourself, w i t h a ruler and compass, by drawing a 3-4-5 t r i a n g l e , and then measuring the angle opposite the longest side with a p r o t r a c t o r . You should bear i n mind, of course, t h a t this check is only approximate. For example, if the angle were r e a l l y 89' 5gf 59", instead of 9 0 ' exactly, you would hardly expect to tell the difference by drawing your figure and then taking a reading with your protractor. Nevertheless, the "Egyptian method" is a sound common-sense method of verifying an experimental fact.
The Egyptians were extremely skillful a t making physical measurements. The edges of the base of the g r e a t pyramid are about 756 feet long; and the lengths of t h e s e four edges agree, with an error of only about two-thirds of an inch. Nobody seems to know, today, how the builders g o t such accuracy. Statement 2 above was not known to the Egyptians; I t was
discovered l a t e r , by t h e Greeks. This second statement is very d i f f e r e n t from the f i r s t . The most iJiroortant difference is t h a t there are i n f i n i t e l y many possibilities f o r a, b and c For instance, you would have to c o n s t r u c t t r i a n g l e s , and take readings with a p r o t r a c t o r , f o r all of the following cases,
.
and so on, endlessly, It seems pretty hopeless to try t o verify o u r general statement by experiment, even approximately. Therefore, a reasonable person would not be. convinced that Statement 2 was true in a l l cases until he had seen some logical reason why it should be true in a l l cases. In f a c t , this is why it was the Greeks, and n o t the Egyptians, who discovered that our second statement is true. The Egyptians had lots of common-sense knowledge of geometry. But t h e Greeks found something b e t t e r , and. much more powerful: they discovered the science of exact geometrical reasoning. By exact reasoning, they learned a great deal that had not been known before t h e i r time. The things that they learned were the f i r s t big step toward modern mathematics, and hence, toward modem science in general.
Problem -S e t 1-1 1.
Try the following experlaent. Take a piece of string, about six feet long, and put it on the f l o o r in t h e form of a loop with the ends free:
Then pull the ends of the string a p a r t , making the loop gradually smaller, and stop when you think that the loop is the size of your o m waist. Then check t h e accuracy of your guess by wrapping the string around your waist. After you have checked, read the remarks at the end of this set of problems.
In this pair of questions, the f i r s t can be answered by IT common sense." State only Its answer. The second requires some arithmetic or algebraic process f o r its s o l u t i o n . Show your work f o r it. a. What I s half of 2? b. VJhat is h a l f of 135,790? Answer as in Problem 2: a. One-third of the distance between two c , i t i e s is 10 miles. What is the entire distance? b. The distance between two c i t i e s is 7 miles more than onethird t h e distance between them. What is the distance
between them?
*I!..
5. 6.
7.
8.
Answer as in Problem 2: a. If a 5-inch piece of wire I s c u t I n t o two p a r t s so t h a t one p a r t is 4 times as long as the o t h e r , what are the lengths of the p a r t s ? b. If a 5-inch piece of wire is c u t i n t o two parts such t h a t a square formed by bending one piece will have four tmes t h e area of a square formed by bending the o t h e r , what are t h e lengths of the parts? If the s i d e s of a triangle are 5, 12, 13, is I t a right triangle? If two s t u d e n t s c a r e f u l l y and Independently measure the width of a classroom with rulers, one measuring from left to r i g h t and t h e other from right to left, they are likely to get different answers. You may check this with an experiment. Which of the following a r e plausible reasons f o r this? a. The rulers have different lengths. b. One person may have lost count of the number of feet In t h e width. Things a r e longer (or s h o r t e r ) from left to right than c. right to left. d. The errors made in changing the position o f the ruler accumulate, and the sum of the small errors makes a discernable e r r o r . 2 S h o w t h a t n - £ + 2 = n i f n = 1. Is the equation true when n = 2? Is it true when n = 3? Is I t always true? a.
b.
If 3 2 , 52 and 7 2 are divided by 4, what is t h e remainder in each case? How many odd integers would you have to square and d i v i d e by 4 t o guarantee t h a t the remainder would always be the same?
Number of points
connected
2
3
Number of regions formed
2
4
10.
8
5
6
16
?
Replace the question mark by the number you think belongs there. Verify y o u r answer by m a k i n g a drawing in which six points on a c i r c l e are connected in all p o s s i b l e ways. The following o p t i c a l i l l u s i o n s show that you cannot always tniafc appearances. 11Things are seldom what they seem; sktm milk masquerades as cream. " From "H.M.s. ina afore" by Gilbert and Sullivan. a . Is CD a continuation of AB? T e s t your answer with a ruler.
A
b.
Are RS and ST equal in length? Compare the lengths with your
ruler or compass. R
c.
Which figure has the greater area?
d.
Which is longer, AB or CD? Check with your ruler.
*11. U s e a ruler to check the accuracy of the measurements of the figure. Show t h a t if these measurements are correct the sum of the areas of the four pieces of the rectangle I s more than the a r e a of the rectangle. Odd, isn't it?
*12.
A trip of
m.p .h.
60 miles is to be done at an average speed of 60
The first 30 miles are done a t 30 m.p. h.
A t what
rate must the remaining 30 miles be covered?
Remarks on Problem 1. Nearly everybody makes a loop about twice as big as it should be. You can get much better results by the following method. The circumference of a c i r c l e is equal to IT times the diameter, and T is approximately equal to 3 . Therefore, t h e diameter is about one-third of the circumference.
If your waist measure I s , say, 21 Inches, th3s means that the loop on the floor should be about 7 Inches across. This will look unbelievably small, but if you have thought the problem out mathematically, you w i l l have the courage of your convictions. This is one of a large number of cases in which even a very crude mathematical approach t o a problem i s b e t t e r than an outright leap in the dark.
1-2.
& Organized Logical Development of Geometry.
If you stop to think, you will realize that by now you know a great many geometrical facts. For example, you know how to find the area of a rectangle, and of a right triangle, and perhaps of a triangle in general, and you know the Pythagorean relation f o r right triangles. Some o f the things that you know are so simple and obvious t h a t it might never occur to you t o even put them i n t o words, l e t alone t o wonder whether or why they are true. The following is a statement of this type: Two straight lines cannot cross each o t h e r in more than one point. But some of them, like the Pythagorean r e l a t i o n , are not obvious a t a l l , but rather surprising. We would l i k e to organize our knowledge of geometry, in an o r d e r l y way, in such a way that these more complicated statements c a n be derived from slmpie statements. This suggests t h a t we ought to be able to make a l i s t of the facts of geometry, with the simplest and easiest statements coming first, and the hard ones coming l a t e r . We might try to arrange the statements in such an order t h a t each statement in the list c a n be derived from the preceding statements by logical re asoning Actually, we shall carry out a program that is very much like this. We w i l l s t a t e definitions, as c l e a r l y and exactly as we can;
.
and we w i l l establish the facts of geometry by giving l o g i c a l p~oofs.
The statements that we prove w i l l be called theorems. (E proving of theomms Ts-not - a spectator-sport, more than -arithmetic is: -t h e best 9 to learn about I t 2s & dolng It. Therefore, in this course, you will have l o t s of opportunities t o
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prove lots of theorems for yourself. ) While nearly a l l of the statements t h a t we make about geometry are going to be proved, there will be some exceptions. The simplest and most fundamental statements will be given without proofs. These statements w i l l be called postulates, and will form the foundation on which we will build. In the same way, we w i l l use the simplest and most fundamental terms of geometry without defining them; these w i l l be c a l l e d t h e undefined terms. The definitions of the other t e r n s that we use will be based on them. A t f i r s t glance, it might seem better to define every term t h a t we use, and t o prove every statement that we make. With a l i t t l e r e f l e c t i o n , we can convince ourselves that this can't be done.
Consider f i r s t the question of the postulates. Most of the time, when we prove a theorem, we do so by showing that It follows l o g i c a l l y from theorems t h a t have already been proved. But it is c l e a r that proofs of theorems cannot always work this way. In particular, the f i r s t theorem that we prove cannot p o s s i b l y be proved t h i s way, because i n this case t h e r e aren't any previously proved theorems. B u t we have to s t a r t somewhere. This means that we have to accept some statements without proof. These unproved statements are the postulates. The purpose of stating postulates I s to make it c l e a r just where we are starting, and Just what s o r t of mathematical objects we are studying. We c a n then build up a solid, organized body of facts about these mathematical objects. Just as we start with some unproved statements, s o we s t a r t with some undefined terms. Most of t h e time, when we g i v e a d e f i n i t i o n of a new geometric term, we define it by m e a n s of o t h e r geometric terms which have already been defined. But i t is c l e a r
t h a t d e f i n i t i o n s cannot always work this way.
In particular, t h e first definition that we state cannot possibly be stated in this way, because in this case there aren't any previously defined geometric terms. But we have t o start somewhere. T h i s means that we Introduce some geometric terms without defining them, and then use these basic terms In our first d e f i n i t i o n s . We s h a l l use t h e
simplest and most; fundamental geometric terms without making any attempt to give definitions for them. Three fundamental undefined terns w i l l be p o i n t , line and p l a n e . Postulates, of course, are not made up at random, (if they were, geometry would be of no I n t e r e s t or importance.) P o s t u l a t e s describe fundamental p r o p e r t i e s of space. In the same way, the m d e f t n e d terns point, l i n e and plane a r e suggested by physical objects. To g e t a reasonably good picture of a p o i n t , you make a dot on paper with a pencil. To g e t a better approximation of the mathematical Idea of a p o i n t , you should f i r s t sharped your p e n c i l . The picture is s t i l l approximate, of course: a dot on paper must cover some area, or you couldn't see it at all. But i f you t h i n k of dots made by sharper and sharper p e n c i l s , you will have a good idea of what we are driving at when we use the undefined term, point. l h e n we use the tern l l n e , we have W m h d the idea of a straight line. A straight line, however, is supposed t o extend infinitely f a r in both directions. Usually, we shall Indicate this in pictures by arrowheads at the ends of the p o r t i o n of the line t h a t we draw, l i k e this:
-
We shall have mot he^ tern, segment, for a figure t h a t looks l i k e
this:
stretched string I s a good approximation of a An even thinner and more tightly stretched string I s a
A thin, tightly
segment.
b e t t e r approximation.
And so on.
Think of a perfectly flat surface, extending Infinitely far in every direction, and you have a good idea of a plane. You should remember that none of the above statements are definitions. They are merely explanations of the I d e a s t h a t people had i n the back of their minds when they wrote the postulates. When we a r e writing proofs t h e Information that we claim to have about p o i n t s , l i n e s and planes will be the information given by the postulates. We have said that theorems are going t o be proved by l o g i c a l reasoning. We have not explained what logical reasoning is, and In fact, we donit know how to e x p l a i n this in advance. As t h e course proceeds, you w i l l get a better and better idea of what l o g i c a l reasoning is, by seeing i t used, and best of a l l by using It yourself. This is the way that all mathematicians have learned to tell what Is a proof and what Isnst. At the beginning of the next chapter, we s h a l l give a short account of the Idea of a set, and a short review of the fundamentals of algebra for real numbers. S e t s and algebra will be used throughout this course, and our study of geometry will largely be based on them. W e s h a l l t h i n k of them, however, as things that we are working with. They w i l l not be an actual p a r t of our system of postulates and theorems. They are supposed to be a v a i l able a t t h e start; some of our postulates w i l l i n v o l v e real numbers; and elementary algebra will be used in p r o o f s . In f a c t , geometry and algebra a r e very closely connected, and b o t h of them a r e easier to learn if the connections between them are brought out as soon as p o s s i b l e .
Problem -Set 1-2 1.
to find the meaning of t h e word "dimension" went t o a dictionary. This dictionary did not give d e f i n i t i o n s as we have them in geometry but d i d .give synonyms of words. He made the following chart. A student wanting
extent
dimension
3.
size
- longest dimension
Point out from the above c h a r t a circular "Mat-, of three terms each of which has its following term a s a synonym. (1n a c i r c u l a r list, t h e first term is assumed to follow the l a s t .) b. Make a c i r c u l a r list which contains f o u r such terms. Make a c h a r t similar t o t h a t i n Problem 1, s t a r t l n p : with some word i n your dictionary. John convinced his mother that he did not track mud onto the living room rug by pointing out t h a t it d i d n o t start raining until 5 o'clock and that he had been in his room studying since 4~30. He mentioned that a person cannot do something if he is not there. The thing he was proving ( t h a t he d i d not track mud) might be regarded as a theorem and the statement about a person not being able t o do something If he I s n o t there might be regarded as a postulate. Make another example of such a convincing argument and point out what corresponds to the theorem, what t o t h e p r o o f , and what to postulates.
a.
*2.
- measurement -
-
13 4.
Janie: Mother:
Janie : Mother:
Janie :
What's an architect? An architect? An architect Is a man who designs buildings. ?4hatfs "designs"? Well--plans. Like we plan a picnic? Yes, quite like t h a t . What are buildings? Oh, Janie, you know -- houses, churches, schools.
Mother: Janie : Mother: Yes, I see. Janie : Consider the above discussion. What were basic undefined terms as far as Janie was concerned? The S t u a r t s have three children. Joe is a senior in high school. Karen is a seventh grader, and Beth I s four. At t h e dinner table : We learned a funny rev word in geometry class Joe : today -- parallelepiped. Karen: What in t h e world Is it? Joe : Well, It's a s o l i d . You know what I mean by a solid figure -- I t t a k e s up some space. And it's bounded by planes. You know what a plane i s , don't you? Beth: Like a windowplane? Joe : The word is a windowpane, b u t thatts the idea. A parallelepiped is a solid bounded by parallelograms. A candy box is one, but I t ' s a special one because the six faces are all rectangles. If you had a candy box and could shove it at one c o m e r you'd get a parallelepiped. Got the idea? In the above discussion what basic, undefined terms did Joe use in his d e s c r i p t i o n ?
6.
*7.
What do you think is wrong with the following faulty
definitions? a. A square is something t h a t is n o t round. b. A right triangle is a triangle each of whose angles has a measure of 90'. c. An equilateral triangle is when a triangle has three sides the s m e length. d. The perimeter of a rectangle is where you f i n d the sum of the lengths of the s i d e s of the rectangle. e . The circumference of a circle is found by m u l t i p l y i n g the diameter by IT. Indicate whether the fol1orri.ng are true or false; a . It is p o s s i b l e to define each geometric term by using simpler geometric terms. b. Exact geometric reasoning leads us to geometric truths t h a t cannot be deduced from measurement. c . Theorems a r e proved only on t h e basis of d e f i n i t i o n s and undefined terms. d . If you are willing to write In a l l the s t e p s , each theorem can be deduced from postulates without making recourse to previous theorems.
Chapter 2 SETS, REAL NUMBERS AND LINES
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2-1. Sets You may not have heard the word s e t used In mathematics bef o r e , but the idea l a a v e r y familiar one. Your family I s a s e t of people, consisting of you, your parents, and y o u r b r o t h e r s and s i s t e r s (if any). These people are t h e members of the s e t . Your geometry class I s a s e t of students; I t s members are you and your classmates. A school a t h l e t i c team l a a s e t of students. A member of a s e t is said t o belong t o the set. For example, you belong to your family and to your geometry c l a s s , and so on. The members of a s e t are o f t e n called i t s elements; the two terms, members and elements, mean exactly t h e same thing. We say t h a t a s e t contains each of I t s elements. F o r example, both your family and your geometry class contain you. If one set contains every element of another set, then we say t h a t the first s e t contains the second, and we say that the second s e t is a subset of t h e f i r s t F o r example, the student body of y o u r s c h o o l cont a i n s your geometry class, and your geometry class I s a subset of the student body. We say t h a t t h e subset l i e s In t h e s e t t h a t c o n t a i n s it. For example, the s e t of a l l violinists lies In the set of a l l musicians. Throughout this book, lines and planes will be regarded as s e t s of points. In fact, a l l the geometric figures t h a t we t a l k about are s e t s o f points. (YOU may regard this, if you l i k e , as a postulate. ) When we say t h a t two sets are equal, or when we write an equality A = B between two sets A and 3, we mean merely t h a t t h e two sets have exactly the same elements. For example, let A be 1 the s e t of all whole numbers between 1 and %, and l e t B toe the 1 1 s e t of all whole numbers between 7 and %. Then A = B, because
-
.
--
each of the s e t a A and 3 has p r e c i s e l y t h e elements 1, 2, 3, 4
and 5. In f a c t , it very o f t e n happens that the same s e t can be described In two different ways; and If t h e descriptions look d i f f e r e n t , this doesn't necessarily mean that the s e t s are different. TWOsets intersect if there are one or more elements t h a t belong to both of them. For example, your family and your geometry class must intersect, because you y o u r s e l f belong t o both of them. But two different classes meeting at t h e same hour do not intersect. The intersection of t w o s e t s l a the set of a l l objects that belong to both of them. For example, the intersection o f the set of all men and t h e set of all musicians is the set of all men musicians. Passing t o mathematical topics, we see that t h e s e t of all odd numbers I s the s e t whoae members are
1, 3, 5 , 7 , 9, 11, 13, 15,
* = *
and so on. The s e t of a l l multiples of 3 I s the s e t whose members are 3, 6, 9, 12, 15, and so on. The intersection of these two s e t s is 3, 9, 15, 21, and s o on; its members are t h e odd multiples of 3. In t h e figure below, each of the two rectangles l a a s e t of points, and their i n t e r s e c t i o n contains exactly two points.
. .
S i m i l a r l y , each of t h e corresponding rectangular regions is a s e t of points, and t h e i r intersection is the small rectangular region in the middle of the f i g u r e . I n t h e next figure, each o f the two lines is a s e t of points, and t h e i r Intersection consists of a
Below, we see two sets of points, each of which is a f l a t rectangular surface. The Intersection of these two s e t s o f points is a part of a s t r a i g h t l i n e .
The union of the two sets 1s the set of all objects that belong to one or both of them. For example, the union of the s e t of a l l men and t h e s e t of a l l women I s the s e t of all adults. The intersection, or the union, of three or more sets is defined
s i m i l a r l y . Thus a triangle is t h e union of three s e t s , each of which i s a subset of a line.
The figure below is the union o f five sets, each of which is a
In some s i t u a t i o n s , it is convenient t o use t h e I d e a of the empty s e t . The empty set Ls the set that h&sno members at a l l . T h i s i d e a may seem a little peculiar at f i r s t , but it I s really v e r y much like the idea of the number 0. For example, t h e following three statements all say the same t h i n g : (1) There are no married bachelors In t h e world. (2) The number of married bachelors In the world I s zero. ( 3 ) The s e t of a l l married bachelors in the world I s the empty set. Once we have Introduced the empty set, then we can speak of t h e Intersection of any two s e t s , remembering t h a t the inters e c t i o n may t u r n out to be the empty s e t . F o r example, t h e Intersection of t h e s e t of a l l odd numbers and the s e t of a l l even numbers is the empty s e t . If you cornmre the definitions of t h e A word of warnine: terms Intersect and Intersection, you w i l l see that these two terms are not related In q u i t e the simple way that you m i g h t exp e c t . When we apeak of the i n t e r s e c t i o n of two sets, we allow
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t h e possibility t h a t the intersection may be empty. But I f we say t h a t the two sets intersect, this always means that t h e y have an
element In common.
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Another word of wamirg: Statements ( 2 ) and ( 3 ) above mean the same thing. B u t t h i s does n o t mean t h a t a set that contains only the number 0 is empty. F o r example, the equation x + 3 = 3 has 0 as I t s only root, and s o the s e t of roots is n o t the empty s e t ; the s e t of roots has exactly one element, namely, the number 0. On the other hand, the set of all roots of t h e equatlon x + 3 = x + 4 really I s the empty'set, because the equation x + 3 = x + 4 has no roots at a l l ,
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1.
Problem Set 2-1 Let A be t h e s e t ( 3 , 5, 6 , 9, 11, 1 2 ) ( t h a t Is, t h e set whose members are 3, 5, 6 , 9, 11, 12) and B be t h e s e t (4,
2.
5 , 7 , 9, 10, 111.
What is t h e i n t e r s e c t i o n of s e t s A and B? What is t h e union of A and B? Consider the f o l l o w i n g sets: S , is t h e set of a l l students In your s c h o o l . Sy is the set of all boys in your student body. S3 is the s e t of a l l girls In your student body. Sj, Is t h e set of all members of the faculty of your school. S is t h e set whose only member I s y o u r s e l f , a student in 5 your school a. Which pairs, of s e t s Intersect? b, Which s e t I s the union of Sn and S ? Which s e t I s the union of S., and S ? c. 5 d. Describe the union of Si and S 4 . e . Which of the sets are sub-sets of Si?
.
In the following figures, consider the l i n e and the c i r c l e as two s e t s of points. In each case, what Is t h e i r i n t e r s e c t i o n ?
3.
Caae I.
4. ,
5.
6. 7.
8.
Case 11.
Case 111.
Consider a s e t of t h r e e boys, ( A , 3, Any set of boys selected f r o m these t h r e e will be called a committee. a . How many different two-member committees can be formed from t h e t h r e e boys? b. Show t h a t any two of t h e committees in (a) intersect. What does the word " Intersect" mean? Consider t h e s e t of all p o s i t i v e even Integers and the s e t of a l l positive odd integers. Describe t h e set which l a t h e union of these t w o s e t s . Describe the Intersection o f the t w o s e t s g i v e n in Problem 5. In the figure, what l a t h e i n t e r s e c t i o n of the triangle ABC and the segment EC? What is t h e i r union?
L e t A be the set of pairs of numbers (m,n) which satisfy the equation 4m + n = 9. Let B be t h e set of pairs of numbers (m,n) which s a t i s f y the equation 2m + n = 5. Find the intersection of the sets A and B.
Let A be the s e t of pairs ( x , y ) for which x + y = 7 . y = 1. L e t B be t h e s e t of p a i r s ( x , y ) for which x What is the intersection o f A and B? 10. Let A be t h e s e t of pairs (x,y) for which x + y = 3 . Let B be the s e t of pairs (x,y) for which 2x + 2y = 7 . What la the Intersection of A and B? 11. Consider the s e t of a l l positive Integers d i v i s i b l e by 2 . Consider the s e t of all p o s i t i v e integers d i v i s i b l e by 3. a . Describe the intersection of these two sets. G i v e i t a f i r s t four members. b. Write an algebraic expression for the Intersection. c. Describe the union of the two sets. Give its first e i g h t members. 12. a . How many s t r a i g h t lines can be d r a w n through 2 points? b. If three points do not lie in a s t r a i g h t l i n e , how many straight l i n e s can be drawn through pairs of t h e points? c. If f o u r points are given and no s e t of three of them lie in a s t r a i g h t l i n e , how many s t r a i g h t lines can be drawn containing s e t s of t w o of t h e points? Answer the same question if five points are given. *d. Answer Question c if n points are given.
9.
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The Real Numbers The f i r s t numbers t h a t you learned about were the "counting numbers" o r " na t u r n 1 numtoe rs" , 2-2.
1, 2, 3, 4, 5, and so on. (YOU knew about these before you learned to read or write. And ancient roan learned to count long before the invention of writing.) The counting numbers never end, because starting w i t h any one of them, w e can always add 1, and get another one. We may think of the counting numbers as arranged on a line, -
4
s t a r t i n g at some point and continuing t o the right, like t h i s :
To t h e left of 1, we put In the number 0, like t h i s :
And the next step I s to put I n the negative whole numbers, l i k e this:
The numbers that we have ao far are c a l l e d the Integers or
-
whole numbers ( p o s i t i v e , negative and zero). The counting numbers are the p o s i t i v e integers, and are o f t e n referred to by t h i s name. Of course, there are l o t s of points of t h e line that have no numbers attached to them so far. Our next step is to put in t h e 1 1 2 1 -T, 2 ¥x-5, 7 and so on. The new numbers fractions 1 3, that we want to put In include a l l numbers t h a t can be expressed as t h e r a t i o 2 of any two integers ( w i t h q not equal t o z e r o ) . 4 We can Indicate a few of these, as samples:
n,
-
-
-
The numbers t h a t we have s o far are called the r a t i o n a l numbers* h his term I s not supposed t o mean that they are in a
better s t a t e of mental h e a l t h than o t h e r and l e s s fortunate numbers. It merely refers t o the f a c t that they are r a t i o s of whole numbers).
The rational numbers form a very large s e t . Between any two uf t h e m there I s a third one; and there are Infinitely many of them between any two whole numbers. It is a fact, however, t h a t t h e rational numbers still do not fill up the line completely. For example, F i s n o t rational; it cannot be expressed as the ratio of any two integers y e t it does correspond t o a p o i n t on the line. ( F o r a proof, see Appendix 111. ) The same I s true :for i/T" and fl,and a l s o f o r such "peculiar1' numbers a s T . Such nonr a t i o n a l numbers are c a l l e d irrational. If we I n s e r t all these extra numbers, in such a way t h a t every p o i n t of the line has a number attached t o it, then we have the r e a l numbers. We I n d i c a t e some samples, l i k e this:
-
You should check that these numbers appear on the scale In approximately the positions where they belong. ( ^/^IS approximately 1 . 4 1 . HOW would you find.,/^?) The real numbers w i l l form part o f the foundation of almost a l l that we are going to do in geometry. And it will be important throughout f o r us to think of the real numbers as being arranged on a l i n e . A number x I s less than a number y if x l i e s t o the left of y.
<
We notice t h a t every negative number lies t o t h e l e f t of every p o s i t i v e number. Therefore, e v e r y negative number is less than e v e r y p o s i t i v e number. For example, -1,000,000 < 1, even though the number -1,000,000 may in a way look "bigger".
We abbreviate t h i s by w r i t i n g x
y.
Expressions of the form x < y are called inequalities. Any inequality can be w r i t t e n in reverse. F o r example, and In general, y > The expression
>
-1,000,000; x means t h a t x < y . 1
x i
y
means that x is less than or equal to y . F o r example, 3 A 5 be5, because 5 = 5. cause 3 < 5, and 5 In your s t u d y of algebra, you have by now learned quite a lot about how the real numbers behave under a d d i t i o n and multiplication. All the algebra that you know can be d e r i v e d f r o m a few t r i v i a l - l o o k i n g statements. These statements are the postu- ' lates f o r addition and multiplication of real numbers. You will find them listed in Appendix 11. You may not have studied algebra on the basis of the postulates; and we are not going to s t a r t such a proceeding now. In t h i s course, we are simply going t o use the methods of elementary algebra, without comment. We should be a little more careful, however, about inequali t i e s and square roots. The r e l a t i o n < defines an order f o r t h e real numbers. The fundamental properties of this o r d e r relation are the following: 0-1, (uniqueness of order) For every x and y , one and only one of t h e following conditions holds: x < y, x = y, x > y. 0-2. (Transitivity of Order) If x < y, and y < z, then x < 2. 0-3. ( ~ d d i t i o nfor ~ n e ~ u a l i t i e s )If x < y, then x + 2 < y + z for every z . (~ultiplicationfor Inequalities) If x < y and z > 0, 0-4. then xz < y z . The statements 0-2. and 0-3. have an Important consequence, which is worth mentioning separately: 0-5. If a < b and x < y , then a + x < b + y .
<
-
T h i s is true f o r the following reason: By 0-3, we know that a + x < b + x and a l s o t h a t b + x < b + y . hat IS, an i n e q u a l i t y is preserved if we add the same number on each side.) By 0-2, these l a s t t w o Inequalities fit together t o
g i v e us a + x < b + y, which is what we wanted. F i n a l l y , we are going to need the following p r o p e r t y of the real numbers : R-1. (Existence of Square Roots.) Every p o s i t i v e number has exactly one positive square root. There is one rather tricky point In connection with square roots. When we say, In words, t h a t x I s a square root of a , t h i s means merely t h a t x 2 = a. F o r example, 3 Is a square root of 9, and - 3 is a square root of 9. But when we write, in symbols, t h a t x =& we mean that x is t h e p o s i t i v e square root of&. Thus, t h e following statements are t r u e or false, as Indicated: True: -3 I s a square root of 9. False: - 3 = i/^". True: fl= 3. False: fl= 3. The reason for t h i s usage is simple, once you t h i n k of it. ~ f c w e r allowed e to denote either the positive root or the negative r o o t , t h e n we would have no way at a l l to write t h e posit i v e square root of 7. ( P u t t i n g a plus sign in front of the express!oni/T'gets us nowhere, because a - p l u s sign never changes the v a l u e of an expression. If y^were negative, then + f l would a l s o be negative),
-
-
-
+
--
1.
Problem S e t 2-2 Indicate whether each of the following l a true or false. a . The real number scale has no end points. b. There exists a point on the real number scale which repreaenta -i/^exactly. 6 The p o i n t corresponding to 7 on the real number scale c. 7 lies between the points corresponding t o and 3. d m Negative numbers are real numbers. Restate the following in words: a. AB < CD. e. 0 < 1 < 2 . b. X > Y . f* 5 1 x > -5c . X Y > YZ. g. x > 0. (3. n 3. Write as an inequality: a. k is a p o s i t i v e number. b. r l a a negative number. c. t Is a number which I s not p o s i t i v e . d. s is a non-negative number. e. g has a value between 2 and 3. f, w has a value between 2 and 3 i n c l u s i v e . g, w has a value between a and b. For which of the following I s I t true t h a t x?
3
2.
<
3.
4.
5.
1/t2-
5
How w o u l d t h e points corresponding to t h e following s e t s of numbers be arranged from l e f t to r i g h t on a number scale in which the positive numbers are to the right of O?
*6. If r and s are real numbers, other than zero, and r
>
s, indicate whether the following are always true (T), sometimes true (s), o r never true (N), a. s < r.
F o l l o w the I n s t r u c t i o n s of Problem 6 for the following:
*7.
The Absolute
Value The idea of the a b s o l u t e value of a number I s easily understood from a few examples: ( 1) The absolute value of 5 I s 5. (2) The absolute value of -5 13 5. (3) The absolute value of TT Is TT. (4) The absolute value of -T is ir, and so on, Graphically speaking, the absolute value of x is simply the distance between 0 and x on the number scale, regardless of whether x lies to t h e left or to the right of 0. The absolute value of x is written as 1x1 2-3.
.
IÑlxlÃ
ÑIÑÑÑI x
-
0
x
*
1ÑÑlX1
ÑÑÑÑÑI 0 X>O
x
The two p o s s i b i l i t i e s for x are indicated In t h e figures. In each of t h e two cases, ] x 1 is the distance between 0 and x .
If a particular number is written down arithmetically, it l a easy t o see how we should write I t s absolute v a l u e .
The reason
l a t h a t In arithmetic, the p o s i t i v e numbers are w r i t t e n as I, 2, 3, 4, and so on. A way to w r i t e negative numbers I s to put minus signs i n front of t h e positive numbers. This gives -1, -2, -3, -4, and so on. Therefore, In arithmetic, If we. want t o " write the absolute value of a negative number, we merely omit the minus
sign, thua, - 1 = 1 - 2 = 2, and so on. We would like t o give an algebraic d e f i n i t i o n f o r 1x1, and we would like the definition to a p p l y both when x is p o s i t i v e and when x is negative. In algebra, of course, the l e t t e r x can represent a negative number. In working algebra problems, you have probably written x = -2 nearly as often as you have w r i t t e n x = 2. If x is negative, then we can't write t h e corresponding positive number by omitting the minus sign, because there i s n t t any minus sign t o omit. There I s a simple device, however, that g e t s around o u r d i f f i c u l t y : if x I s negative, then t h e correspondi n g p o s i t i v e number I s -x. Here are some examples: x = -1, -x = -(-1) = 1; that l a , if x a -1, then -x = 1. x = -2, -x = -(-2) = 2; that is, if x = -2, then -x = 2. x = -5, - x = -(-5) = 5; that la, If x = - 5 , then -x = 5. In each of these cases, x I s n e g a t i v e and - x la the corresponding p o s i t i v e number. And In f a c t , this l a what always happens. Since we knew a l l along t h a t 1x1 = x when x Is positive or zero, It follows that the absolute value is described by the following t w o statements: (1) If x I s positive or zero, then 1x1 = x, (2) If x is negative, then 1x1 = - x . If t h i s s t i l l looks doubtful to you, try substituting various numbers f o r x. No matter what number x you pick, one of the cond i t i o n s above w i l l apply, and will give you the right answer f o r t h e absolute value.
Problem Set 2-3 1. Indicate which of the following are always true: a. 1-31 = 3 . b * 131 = -3. C. 1 2 - 71 = 1 7 - 2 1 , d. 1 0 - 51 = 1 5 - 01. e. i n = n . * 2 . Indicate which of the following are always true:
,
+
!dl
1 = I d + 11. Complete these statements: a. If 0 < r, then = b. If 0 > r, then lrl =-. c. If 0 = r , t h e n r ! = T h e following three examples give a geometric Interpretation t o algebraic statements. x < 2. e.
3.
-.
4.
I
All p o i n t s of the scale t o the l e f t of 2. I
-1
-2
1 The -
-3
T w o
I
-2
0
I
-1
-
2
1
s e t of points between 2 and -2.
A
f
I
i
I
1
-
-
p o i n t s .
-
I
I
1
f
0
1
2
3
[ s e c . 2-31
Continue as above f o r the following problems: a. x < 0. e. 1x1 = 1. b. x = 1. f. Ixl 1. g. 1x1 > 1. d , x A 1. h. 1x1 2 0 . 5. a. How would the set of points represented by x 0 d i f f e r from the s e t represented by x -> O? b. HOW +would t h e s e t of points represented by 0 <Â x <' 1 differ from the s e t represented by 0 <- x <- I?
-
2-4. Measurement of Distance The f i r s t step in measuring the distance between two points P and Q la to l a y down a ruler between them, like thiat P
0 RULER
Of course we want to use a straight ruler, since we cannot expect to get consistent results if o u r rulers are curved or notched. A s t r a i g h t r u l e r has the property that however it is placed between P and Q the line drawn along I t s edge I s always the same. In other words, this line Is completely determined by the two given p o i n t s . We express this basic property of lines as our f i r s t geometric postulate: Postulate 1. Given any two d i f f e r e n t points, there Is exactly one line which c o n t a i n s both of them. We shall o f t e n refer to t h i s postulate, b r i e f l y , by saying that every two points determine a line. This Is simply an abbreviated way of stating Postulate 1. To designate the l i n e determined by two points P and Q we use the notation (The double arrow will recall our picture of the line. ) Of course we can always a b b r e v i a t e by I n t r o d u c i n g
s.
a new l e t t e r and calling t h e line L, or W, or a n y t h i n g else. Now let us consider the marks on the ruler* and the actual
diatance between P and Q. The easiest way to measure the distance is t o place t h e ruler l i k e t h i s : P Q
This gives 7". Of course, there I s no need t o put one end of the ruler at P. We m i g h t put it like this: P Q
In t h i s case, the distance between P and Q, measured in Inches, la 9 2 = 7, as before. On many rulers that are sold now, one edge I s l a i d o f f in inches, and the other edge In centimeters. Using the centimeter scale, we can measure the distance between P and Q like t h i s :
-
This gives t h e distance as approximately 18 cm., where cm. stands f o r centimeters. A foot I s , of course, 12", and a yard I s 36". A meter I s a hundred centimeters; m, stands f o r meters* A millimeter I s a t e n t h of a centimeter ( or =ofi a meter) ; m. stands for millimeters. We can therefore measure the distance between
P and Q I n at least t h i s many ways: 7 in., ft-, yds-
&
18 em., 180 mm.,
.18 m , ,
T h a t I s , t h e number we get, as a measure of t h e distance, depends on the u n i t of measure. We can use any unit we like, as long as we use it consistently, and as long as we say what unit we are us ing
--
.
--
1.
Problem S e t 2-4 What common f r a c t i o n s (or integers) are needed to complete the following table? ft. = yd. a. 2 in. =
-yd.
-
b.
=. 2.
-
- in. -ft. =
4.
5.
=
2 3
yd.
What numbers are needed to complete the following table? a. 500 mm. = cm. = m. b. mrn. = 32.5 cm. = m. mm. = cm. = 7 . 3 2 m. a. Suppose you d e c i d e to use the w i d t h of an 8 1 in. by 11 in. sheet of paper as a u n i t of length. What I s the length and the w i d t h of the sheet in terms of t h i s u n i t ? b. Repeat the problem with the length of the sheet as your new u n i t . I f the lengths of the sides of a t r i a n g l e are 3 ft., 4 ft., and 5 ft., I t Is a right triangle because 32 + 4 = 5 2 , V e r i f y t h a t t h e Pythagorean relationship still h o l d s i f the lengths above are expressed I n Inches. If the length of each side of a square is 4 ft. I t s perimeter is 16 ft. and I t s area I s 16 sq. ft. Observe that the
-
3.
-,
In. = 4 1 ft. =
-
-
-
-
numerical value of the perimeter is equal to t h e numerical value of the area. a. Show that the numerical values of the perimeter and area w i l l no longer be equal t o each o t h e r if t h e length of t h e side is expressed In Inches.
*6.
b, In yards. Generalize Problem 4. Given that the numbers a, b and c are the number of u n i t s I n t h e sides of a triangle If some p a r t i c u l a r u n i t of length I s used and t h a t a 2 + b2 = c 2 Show that t h e Pythagorean r e l a t i o n s h i p k i l l s t i l l h o l d If the u n i t of l e n g t h i s multiplied by n. (Hint: The lengths of If a, b and c seem too the sides will become , ;a Ãb and c abstract use 3, 4 and 5 at f i r s t . ) Generalize Problem 5. Show that if the numerical values of the area and p e r i m e t e r of a square are equal for some particular unit of measure, t h e n they w i l l not be equal for any other unit. (Hint: Start by letting t h e number s be t h e length of the s i d e of the square f o r some u n i t and equating the area and p e r i m e t e r formulas.)
.
.
*7.
----
Choice of a Unit of Distance We have noticed t h a t t h e choice of a unit of distance I s merely a matter of convenience. Logically speaking, one u n i t works as well as another, for measuring distances. Let us theref o r e choose a unit, and agree t o talk i n terms of t h i s unit In a l l of our theorems, (it w i l l do no harm to think of our u n i t aa being anything we like. If you happen to like inches, feet, yards, centimeters, cubits, or furlongs, you are free to consider t h a t these are the u n i t s t h a t we are using. A l l of our theorems will hold t r u e for 9 unit.) Thus, to every pair of points, P, Q there w i l l correspond a number which l a t h e measure of the distance between P and Q I n terms of our u n i t . Such numbers will be used e x t e n s i v e l y In o u r work, and it would be very inconvenient to have to be continually repeating the long phrase "measure of t h e distance between P and Q in terms of o u r u n i t " . We shall therefore s h o r t e n t h i s phrase to "distance between P and Q", trusting that you will be able to fill I n the remaining words if I t should ever be necessary.
2-5.
---- -
We can now describe t h i s s i t u a t i o n in the follow^ precise fo m : Postulate 2. h he Distance Postulate. ) To every 1 pair of different points t h e r e corresponds a unique positive^ number.
I 1
1 1
Definition. The distance between two points l a the p o s i t i v e number given by the Distance Postulate. If the points are P and Q, then the distance I s denoted by PQ. It w i l l sometimes be convenient t o allow the possibility P = Q, that is, P and Q are the same point; in t h i s case, of course, the distance is equal to zero. Notice that distance I s d e f i n e d simply for a pair of points, and does not depend on the order in which the points are mentioned. Therefore PQ is always the same as QP. Some of the problems you w i l l be asked to do w i l l Involve various units of distance, such as feet, miles, meters, e t c . As noted above, o u r theorems will be applicable to any of these u n i t s , p ~ o v i d e dyou consistently yse J u s t one u n i t throughout any one theorem. You can use Inches in one theorem and f e e t In a n o t h e r , If you wish, but not both in the same theorem.
2-6.
An I n f i n i t e
Ruler
A t the beginning , o f this
chapter we l a i d o f f a number-scale
on a line, like t h i s :
We could, of course, have compressed the scale, like t h i s :
or s t r e t c h e d
It, l i k e t h i s :
But l e t us agree, l a y off on a line labeled x lies at example, consider numbers 0, 2, -2,
f r o m now on, that every number-scale t h a t we is to be chosen In such a way t h a t the p o i n t a distance 1x1 from the p o i n t labeled 0. For the points P, Q, R, S, and T, labeled with the -3, and 4, as in the figure below:
Then PQ = 2, PR = 2, PS = 3 and PT = 4. If we examine various pairs of points on the number-scale, it seems reasonable to f i n d the distance between two points by t a k i n g t h e difference of the corresponding numbers. F o r
36 example,
PQ = 2,
and 2 Q T = 2 , and 2 S Q = 5, and 5 RT = 6, and 6 Notice, however, that if we look at order, and perform the subtractions
= 2
-
0; 2;
= 2 - (-3);
3
4
-
= 4 (-2). the pairs of points in reverse In reverse order, we w i l l get the wrong answer e v e r y time: instead of g e t t i n g the distance (which I s always p o s i t i v e ) , we w i l l g e t the corresponding negative number. This d i f f i c u l t y , however, is easy to g e t around. A i l we need t o do I s t o take the absolute value of the difference of t h e numbers. If we do t h i s , then a l l of o u r positive right answers will still be right, and a l l of our negative wrong answers will become right. Thus we aee that t h e distance between two p i n t s IEIthe absolute value of the difference of the corresponding numbers. Surety all thia seems reasonable. But s u r e l y we have not proved it on the basis of the only postulatesthat we have written down so far. (And, In f a c t , it cannot be proved on the basis of the Distance Postulate.) We therefore aum up the above discussion In the form of a new postulate, like this:
-
-----
--
- 7 -
I
-
Postulate 3 . h he Ruler Postulate. ) The p o i n t s of a line can be placed i n correspondence with the real numbers In such a nay that (1) To every point of the line there corresponds exactly one real number, ( 2 ) To every real number there corresponds e x a c t l y one point of the line, and ( 3 ) The distance between two points Is the absolute value of t h e difference of the corresponding
-
numbers
.
We c a l l t h i s t h e Ruler P o s t u l a t e because In e f f e c t it f u r nishes us with an I n f i n i t e ruler, w i t h a number-scale on It, with w h i c h we can measure distances on any line.
Definitions. A correspondence of t h e s o r t d e s c r i b e d In P o s t u l a t e 3 is called a coordinate system f o r the l i n e , T h e number corresponding to a given point I s c a l l e d t h e coordinate of the point. Problem Set 2-6 1. Simplify: a. 1 3 - 61. d. 1-4 (-2)l. b. 1 6 - 31. e. la (-a)l. c. 1-2 - 1 f. J a l I-a(. 2. Using the kind of coordinate system d i s c u s s e d in the text, f i n d the distance between p o i n t pairs w i t h the following coordinates : a. 0 and 12. f. -5.1 and 5.1. b. 12 and 0. g. -/"? and 0. c. 0 a n d -12. h . x, and x2. d. -12 and 0. i. 2a and -2a. 1 e. - 3 3 and - 5 . j. r s and r + s . 3.
--
-
-
The lower numbering on this scale was put there by Jim. Pete began the upper numbering but q u i t . a. Copy the scale and write in the rest of Pete I s numbering. b. Show how to f i n d the distance from P to Q, f i r s t by using Jim's scale and then by using Pete's scale.
Do the same f o r the distance from ^f to P. Suppose in measuring the distance between two points P and Q you intended to place t h e zero of t h e number-scale at P and read a p o s i t i v e value at Q. However, you happen to place the 1 and Q is farther to t h e r i g h t . number-scale so that P is at -nc.
4.
*5.
*6.
7.
How is it still possible to measure the distance PQ? Consider a c o o r d i n a t e system of a line. Suppose 2 I s added t o the coordinate of each point and t h i s new sum 13 assigned to the point. a. W i l l each point then correspond to a number and each number t o a point? to. If two p o i n t s of t h e line had coordinates p and q in the coordinate system given, what numbers are assigned to them In the new numbering? c. Show t h a t the formula (Number assigned to one point)-(Plumber assigned t o " o t h e r point) [ g i v e s the d i s t a n c e between the two points. d . Does the new correspondence between points and numbers satisfy each o f the three conditions of Postulate 3? (if i t does it may- be called a coordinate system.) Suppose a coordinate system is s e t up on a l i n e 3 0 t h a t each p o i n t P corresponds to a real number n, If we replace each n by -n, then the point P will correspond to a number -n. Show t h a t t h i s correspondence Is a l s o a coordinate system f o r t h e line. ( H I N T ; It is apparent that each point will have a number associated w i t h I t and each number a p o i n t . You must show In addition that the absolute value of the difference of the numbers assigned to the two points w i l l , remain unchanged when the numbering l a changed. ) In a c e r t a i n county the towns of Alpha, Beta and Gamma are colllnear ( on a l i n e ) but n o t necessarily In that order. It is 16 miles from Alpha to Beta and 25 miles from Beta to Gamma. a . Is it possible to tell which town l a between t h e o t h e r two? Which town l a not between the other two? b. There m i g h t be t w o different values f o r the distance from Alpha to Gamma. Use a sketch to determine what these are.
-
If you are g i v e n the a d d i t i o n a l information that t h e d i s tance from Alpha t o Gamma i s 9 miles, t h e n which town is between the other two? d . If t h e distance between Alpha and Beta were r miles, the distance from Alpha t o Gamma s miles, and t h e d i s t a n c e from B e t a t o Gamma r + s miles, which city would be between the other two? A, B, C are three c o l l l n e a r points. A and B are 10" apart, and C I s 15" from B. Is there just one way to arrange these po i n t s ? Explain. Three different coordinate s y s t e m s are assigned t o the same line. T h r e e f i x e d points A, B, C of the l i n e are assigned values as follows: With the f i r s t system the coordinate of A Is -6 and t h a t of B I s -2. With the second t h e coordinates of A and C are 4 c.
and -3 r e s p e c t i v e l y . With the third the respective
coordinates of C and B
are 7 and 4. W h a t point l a between the o t h e r two?
Evaluate AB
+
BC
+
AC.
-
The R u l e r Placement Postulate Betweenness- Segments And Rays The Ruler P o s t u l a t e ( P o s t u l a t e 3) t e l l s us t h a t on any line, we can set up a coordinate system by l a y i n g o f f a number-scale. Thia can be done i n l o t s o f d i f f e r e n t ways. For example, given a p o i n t P o f the l i n e , we can s t a r t by making P t h e z e r o - p o i n t . And we can then l a y o f f t h e scale In either d i r e c t i o n , like t h i s : P
2-7
-
-
t
-3
I
-2
I
-1
I
0
I
1
I
1
2
3
This means that given another point Q of the line, we can always choose the coordinate system in such a way t h a t Q corresponds t o a p o s i t i v e number, like t h i s : P Q
-3
I
I
1
I
0
1
2
3
I
I
f
-2
-1
1
or t h i s :
L e t us w r i t e t h i s down, for future reference, in the form of a
postulate
.
he Ruler Placement Postulate 4. Given t w o points P and Q of a line, the system can be chosen in such a way that of P is z e r o and t h e coordinate of Q is
Postulate.) coordinate the coordinate positive.
Everybody knows what I t means to say that a point B I s between two points A and C. It means that A, B and C l i e on t h e same line, and that they are arranged on t h e line like t h i s : #
A
I I
B
I I
C
or like this:
If we are going to use betweenness as a mathematical Idea, however, we had better give a mathematical definition t h a t s t a t e 8 exactly what we mean, because t h e feelings that we have In o u r bones are not necessarily reliable. To see this, let us look at the corresponding situation on a circle. In the figure on the left,
it seems reasonable to say that B is between A and C. But C can be moved around the c i r c l e In easy stages, without passing over A or B, so as to l i e J u s t to the left of A, as in the righthand figure. In the final position, I n d i c a t e d by the exclamation point, It looka as if A is between B and C . In t h i s respect, circles are tricky. Given any three points of a circle, I t I s quite reasonable t o consider t h a t each of them I s between t h e other two. Betweenness on a line is n o t at a l l tricky. It I s easy to say exactly what it means for one point of a l i n e to be between two o t h e r s . We can do t h i s In the following way: Definition. 3 l a between A and C If (1) A, 3 and C are distinct points on the same line and (2) AB -I- BC = AC. It is easy to check that t h i s d e f i n i t i o n really expresses our common-sense I d e a of w h a t betweenness ought t o mean. It may be a good idea t o explain, however, t h e way In which language is ordinarily used In mathematical definitions. In the d e f i n i t i o n of betweenness, two statements are connected by the word if. What we really mean I s that the statements before and after the word If are completely equivalent. Whenever, In some theorem, or problem, we are given or can prove that conditions (1) and ( 2 ) both hold, then we can conclude that 3 l a between A and C. And whenever we find that B is between A and C then we can conclude that (1) and ( 2 ) both hold. T h i s I s not a s t r i c t l y l o g i c a l use of the word if, and parthular the word never uaed In t h i s way In postulates, theorems or problems. In definitions, however, i t l a common. The following theorem describes betweenness in terms of
-
-
coordinates on a line.
-
Theorem 2-1. L e t A, B, C be three points of a l i n e , with coordinates x, y, z . If X < Y < z ,
then B is between A and C. Proof: Since x < y < z, we know t h a t the numbers y x, z y, and z x are a l l p o s i t i v e . Therefore, by d e f i n i t i o n of t h e absolute value,
-
-
-
Therefore, by the Ruler Postulate, AB = y
-
X,
m = z - y , AC 3 z X.
-
Theref o r e AB+
BC = ( y
-
- x +
X)
+
( Z
- y)
2
= z - x = AC.
T h e r e f o r e , by the definition of betweemeas, B i s between A and C, which was to be proved.
-
-.-Problem Set 2-78 A number-scale I s placed on a line with -3 falling at R and 4 at S. - If t h e Ruler Placement Postulate is applied with 0 placed on R and a positive number on S , what will this number be? Same question 1f -4 f a l l s at R and -10 a t S. Same question if 8 f a l l s at R and -2 at S. Same question if -4 f a l l s at R a n d 4 at S. Same question if 5.2 f a l l s a t R and 6.1 a t S. Same question if x-, f a l l s a t R and 5 at S . Explain b r i e f l y how t h e Ruler Placement Postulate simplifies the procedure given by the Ruler Postulate f o r computation of distance between two points,
Ã
2.
Suppose R, 3 and T are collinear points. What must be true of the lengths RS, ST and RT If S is to be between R and T? ( See d e f i n i t i o n of between. ) A
C
B
BC each equals 8. The coordinate of C is 6. The coordinate of B I s greater than t h e coordinate of C. What are t h e coordinates of A and B? If a, b and c are coordinates o f collinear points, and If bl, what I s the coordinate of t h e a c l + l c - b) = la p o i n t which l i e s between t h e other two? Be able to juatify y o u r answer. If x,, x2 and x3 are coordinates of points on a l i n e such that x3 > x, and x2 < xl, which point is between the other two? Which theorem would be used to prove your answer? AC and
-
-
Consider a coordinate system in which A is assigned t h e number 0, B is assigned the positive number r y E the number 1 2 r, and F the number 3 r. Prove that: a. A E = E P = F B b. E is between A and F. Prove: If A, B and C are three points of a line with coordinates x, y and z r e s p e c t i v e l y and If x > y > z, t h e n B is between A and C.
Theorem=. Of any three d i f f e r e n t points on the same line, one I s between the o t h e r two. Proof: L e t t h e points be A, B and C. By the Ruler Postulatey there I s a coordinate system f o r the l i n e . L e t the coordinates of A, B, and C be x, y, and z . There are now six possibilities:
(1) x (2) x
< Y < 2, < < Y, < x < 2, < < x,
(3) Y (4) Y (5) Z < X < Y , (6) < Y < x. In each of these cases, Theorem 2-2 f o l l o w s by Theorem 2-1. In cases (1) and (6), B l a between A and C. In cases (2) and f 4 ) , C Is-between A and B. In cases ( 3) and ( 5 ) , A I s between B and C.
u,
Of t h r e e d i f f e r e n t polntg on the a m 0 lac, o n l y Theorem one la between the other two. Restatement. If A, B and C are three different points on the same line, and 3 l a between A and C , then A I s not between B and C, and C is not between A and B. I t o f t e n happens t h a t a theorem Is easier t o read, and easier to refer to, If I t l a s t a t e d In words. But t o prove theorems, we usually need t o s e t up a notation, g i v i n g names to the objects that we w i l l be talking about. F o r t h i s reason, we shall o f t e n give restatements of theorems, I n the s t y l e t h a t we have just used f o r Theorem 2-3. The restatement gives us a s o r t of head-start In the proof. ) Proof: If 3 l a between A and C, then AB + BC = AC. , If A Is between B and C, t h e n BA + AC = BC. What we need to prove is t h a t these two equations cannot both hold at t h e same time. If the f i r s t equation h o l d s , then AC EC = AB. If the second equation holds, then AC BC = -BA e -AB. Now A 3 is positive, and -AB is negative. Therefore, these equations cannot both be true, because the number AC BC cannot be both positive and negative. [ s e c . 2-71
-
-
-
In an e n t i r e l y similar manner we can show t h a t C is not between A and B. Definitions. For any two points A and B the segment is t h e s e t whose p o i n t s are A and B, together with a l l p o i n t s that are between A and 3. The points A and B are called the end-
-
- AB.
points of Notice that there is a big difference between the segment and the d i s t a n c e AB. The segment l a a geometrical figure, that Is, a s e t of points. The d i s t a n c e is a number, which t e l l s us how far A Is from 3. Deflnltion. The distance AB is called t h e l e n g t h of the segment A ray Is a f i g u r e that looks like t h i s :
m.
A B The arrow-head on the right Is meant t o Indicate t h a t the ray includes all points on the l i n e t o the r i g h t of the point A, plus Notice t h a t when the point A I t s e l f . The ray la denoted by
-
a.
a,
we write we simply mean the ray that starts at A , goes through 3, and then goes on In the same direction forever. The ray might
look l i k e any of ths following:
1,
That is, t h e arrow In the symbol AB always goes from left to right, regardless of how the ray is pointed In space. Having explained informally what we are d r i v i n g at, we proceed t o give an exact definition.
Definltlona. L e t 4 and 3 be points of a line L, The 2 Is the set which is t h e union of ( 1) t h e segment and (2) the s e t of all points C for which It Is true t h a t B i g between A and 0. The p o i n t A Is called the end-point of f i . These two parts of the ray are as i n d i c a t e d : . a
yy
(I) '-Sf
/-Â
0
A
C
If A is between B and C on L, then the two r a y s ' ' g o in opposite direction, like t h i s :
-
3
AB
-
-
-
\_/^
-
A
C
A^ and &S
6
8
are Beflnltton. If A I s between B and C, then ($ and c a l l e d opposite mys, Note t h a t a pair of points A, B determines six geoometrlc figures :
0
The line A3, The segment The ray
E,
¥ÑÑÑÑÃ
&, -Ã
The ray BA, The ray opposite to
3,
-
-È
The ray opposite to BA. The Ruler Placement Postulate has three more simple and
useful consequences. Theorem 2-4, h he Point Plotting heo or em) L e t be a ray, and let x be a p o s i t i v e number. Then t h e r e Is exactly one -> p o i n t P of AB such that A P = x . Proof: By the Ruler Placement Postulate, we can choose the coordinate system on the line In such a way that the coordinate of A I s equal to 0 and the coordinate of B Is a p o s i t i v e number r: A B P -
SB
-----------
0
1
r
r
x
(In the figure, the labels above the line represent points, and the labels below the l i n e represent the corresponding numbers.) Let P be the p o i n t whose coordinate I s x. Then P belongs to i6, and AP = !x 01 = 1x1 = x, because x la positive. Since only one point of the ray has coordinate equal t o x, only one point of the ray l i e s at a distance x from A. Definition. A point B I s called a mid-point of a segment AC if B l a between A and C, and AB = BC.
-
-
T h e o r e m s . Every segment has exactly one mld-point.. Proof, On the segment we want a p o i n t B such t h a t AB = BC. We know, by definition of a segment, that 3 is between A and C. Therefore, AB BC = AC. From these two equations we conclude 1 AC. Since B is to lie on segment AC t h a t 2AB = AC, or AB a .2 it must a l s o lie on ray AC, and Theorem 2-4 t e l l s us t h a t there l a exactly one such point B. Definition. The mid-point of a segment 1s s a i d to bisect the segment. More generally, any figure whose Intersection w i t h . a segment is t h e mid-point of the segment I s s a i d t o bisect the
+
-
segment.
-
1. 2.
Problem S e t 2-7b If three p o i n t s are on a line, how many of them are not between the other two?
Each of the following la a particular case o f what definition or theorem? If three colllnear points R, S and T have coordinates respectively 4, 5 and 8: a. S I s between R and T because 4 < 5 and 5 < 8, b. R cannot be between S and T a i n c e S I s between c.
R and T. S I s between R and T because RS
+
ST = RT.
4
3,
*4.
Describe in mathematical language what points are included In: 4 a. XY b. XY Show that the restriction "between A and C" in the d e f i n i t i o n of the midpoint of AC is unnecessary by proving t h e following theorem: If B i a any point on the line such that AB = BC, then B 3.5 between A and C. ( ~ i n t : Show t h a t A cannot be between B and C nor 0 between A and 3. Use algebra I n showing this. Use Theorem 2-2 to f i n i s h the proof.)
8
*5.
*6. 7.
Suppose t h a t P is a point on a line M and r I s a p o s i t i v e number. Which of the previous theorems shows that there are exactly two points onM whose distance from P l a the given number r? Prove that If B is between A and C, then AC > AB. a . Copy t h e following paragraph. Supply the appropriate
missing symbol, if any, over each l e t t e r pair. XZ contains pointa Y and R, but XZ contains neither points Y nor R. . R belongs to XZ but Y does not.
YZ b.
+
ZR = YR.
Make a d r a w i n g showing the relative position of t h e four points.
Review Problems 1.
Consider the following sets: S, 18 the set of a l l boys in t h e 10th grade. Sg is the set of all girls I n the 10th grade. 3, I s the s e t of a l l 10th grade geometry students. SL Is the set of all students In high school. S I s the set of all 10th grade students.
5
a. b. c.
What l a the intersection of S,and S ? 5 W h a t I s the union of Sa and 3h? What l a the Intersection of Sg and Sh?
d.
e. 2.
a. b. c
3.
.
a.
b.
4.
a. b. c.
and Sg? What is t h e intersection of S-, and Sp? How many squares does a given p o s i t i v e number have? How many square r o o t s ? Is i/^ever negative? Draw a line and locate t h e following points on It. (The coordinate of each point is given in parentheses. ) Use any unit of measure you choose, but use the same unit t h r o u g h o u t . P (21, Q (-11, R (01, s ( - 3 1 , T ( 4 ) . Find PQ, RT, TR, FT, QS. If a > b, then a - b is If 0 < k and k^ < 4, then k I s If a < b then a b is
What
i s t h e union of S.,
-
-
.
.
Write a n equation t h a t describes the relative positions of these three points. b. Under what c o n d i t i o n would 3 be the m i d p o i n t of Pour p o i n t s A, 3, C, D are arranged along a l i n e so t h a t AC > AB and BD < BC. Picture t h e line with the four points in place. Is t h e r e more than one possible order? Explain. The l e t t e r pairs contained i n the following paragraph are either numbers, lines, line segments, or rays. Indicate which each l a by placing the proper missing symbol, If any, above each letter pair. 'AB + BC = AC. DB contains points A and C, but DB containa neither point A nor point C, A belongs to DB but C does not." D r a w a picture t h a t i l l u s t r a t e s your response , A Is t h e s e t of a l l integers x and y whose sum is 13. B l a the s e t of a l l integers whose difference is 5 . What is t h e Intersection of A and B? a.
z?
6
7.
8.
9.
10.
11.
John s a i d , "My house I s on West Street halfway between B f l l * ~ house and Joe's house." P e t e said, "So l a mine!" What can you conclude concerning John, and Pete? N men a i t on a straight bench. Of how many may I t be said, 'He a i t s between two people?" Use the figure below t o answer questlona a. through e.:
Describe the intersection of triangle AEP and rectangle ABCD. b. Describe the intersection of segment and rectangle ABCD. Describe the union of segments AF, EP, and E . c. d. Describe the Intersection of segments and E. e. Describe the union of triangle AEP and segment Given a group of five men (Meaara Andrews, Brown, Crawford, Douglas, and ~ v a n s ) a. ~ r o mt h e group, how many different 4-man committees can be formed? b. 2-man? c. 3-man? Given that A, B and C are c o l l i n e a r and that A 3 = 3 and EC = 10, can AC = 6? U v e a picture to explain your answer. Indicate which of the following statements are true and which are f a l s e . For any t h a t are falae, give a correct
a.
--
12.
13. 14.
.
.
z.
Looking at t h i s number-scale, Jack said, h he l e n g t h of RQ is 1 x-y 1 .'I S a m maintained that when giving the length of it would be Just as correct to use simply y-x. Do you agree with Sam? Explain. 16. The first numbering of the points on the line below represents a coordinate system. Which of t h e other numberings are not coordinate systems according to Postulates 2 and 3? - . . -3 -2 -1 1 2 3 4 5 6 7 -3 -2 -1 o 1 2 3 a. -7 -6 -5 -4 3 4 5 4 3 2 1 0 b. 0 1 2 12 13 14 15 16 17 18 19 20 21 c. 11 d. -11 -12 -13 -14 -15 -16 -17 -18 -19 -20 -21 e. -3 -2 1 0 -1 2 3 4 5 6 7 *17. Consider the points of a line whose coordinates are d e s c r i b e d as follows: e. x = - 3 . a* x < 3. b. x - 1 . f. 1x1 3. c. x > 2. 6. 1x1 > 2. d. x 1. h. I x l > O . -7
0
<
<
Whlch of the above sets l a a ray?
A segment?
A polnt?
A line?
Chapter 3 LIKES, PLANES AND SEPARATION
3-1.
Lines and Planes in Space, --
In t h e l a s t chapter, we were t a l k i n g only about l i n e s and t h e measurement o f distance. We shall now proceed t o the study of planes and space. We recall -that our basic undefined terms are point, line and plane. Every line I s a s e t of p o i n t s , and e v e ~ y plane Is a s e t of p o i n t s . D e f i n i t i o n . The s e t of all p o i n t s Ts c a l l e d space. In t h i s section we will explain some of the terms we are going t o use in talking about points, lines and planes, and s t a t e some of t h e basic facts about them. Most o f these basic facts w i l l be stated as p o s t u l a t e s . Some of t h e m w i l l be s t a t e d as theorems. These theorems w i l l b e so simple that it would be reasonable t o accept them w i t h o u t p r o o f , and call them postulates. We do not do this, however; the first of them is going to be proved In t h i s sectlon, and the r e s t of them will be proved, on the basis o f t h e postulates, in a later chapter. For the present, however, let us not worry about this question, one way or another; l e t us simply t r y t o g e t these basic facts straight.
--
Problem Set 3-la
a piece of paper, or on t h e blackboard, place two marks to represent p o i n t s A and B. How many d i f f e r e n t lines can you draw through both A and B? What happens i f you consider "linel1 In a sense other than "straight"? 2. Take a piece of s t i f f cardboard or your book. Can you support it in a fixed position on the ends of two pencils? What is the minimum number of p e n c i l s needed to support it in this way? 3. Think of one cover of your book as p a r t of a plane. How many points are needed to determine t h i s plane? 4. How many end-points does a line have? How many end-points does a line segment have? 1.
On
Definition, A set of p o i n t s is which contains all the p o i n t s of the Definition, A set of points is which contains a l l t h e points of the
colllnear if there Is a line set.
coplanar if there I s a plane set,
For example, In the above figure of a triangular pyramid, A, E and B are colllnear, and A, F and C are colllnear, but A, B and C are not colllnear. A, B, C and E are coplanar, and A , C, D, F and G are coplanar, but A , B, C and D a r e not coplanar. One of t h e properties we d e s i r e f o r the s e t s of points which we c a l l lines, planes and space is that they should contain lots of points Also, a plane should in some sense tie "biggertf t h a n a l i n e and space should be ' b i g g e r than any plane. The existence of plenty of p o i n t s on a l i n e is insured by the Ruler Postulate; for planes and space the following postulate w i l l give us the properties we want:
P o s t u l a t e 5. ( a ) Every plane c o n t a i n s a t least three non-colllnear points. (b) Space contains at least four non-coplanar points, For convenience in reference we repeat Postulate 1. Given any two different points, there Is exactly one l i n e which contains t h e m .
Postulate 1.
Theorem 3-1, Two different lines Intersect in a t most one point.
The proof of t h i s follows from Postulate 1. It is Impossible for two different lines to intersect in two different p o i n t s F and Q because by P o s t u l a t e 1 there is only one l i n e t h a t contains P and
-
Q
--
Problem S e t 3 - l b
Li and Lp are different lines. 2. P o i n t P lies on L, and Lp. 3 . Point Q lies on L, and Lp. What can you say must be true about P and Q? 2. How many lines can contain one given point? two given points? any three given points? ¥ÇÑ> w 3. The diagram shows three different lines AB, CD, and EF, whose ¥Ç view is partially obstructed by a barn. If ^B and CD intersect t o t h e left of t h e barn, which postulate says t h a t they cannot a l s o intersect to the right of t h e barn? 1. Given:
11.
1.
Draw a diagram to illustrate each part of t h i s problem and just i f y your answers in terms of Postulate 1. a . How many l i n e s can be drawn through both of two f i x e d points?
How many lines can be drawn through three points taken t w o at a t i m e ? a. How many lines can be drawn through four coplanar points, taken two at a time, if no t h r e e of the points are collinear? ( ~ i n: t Call the points A, 5, C, D. ) b. How many lines would there be if p o i n t s A , B, and C were collinear? c . Draw a diagram f o r (a) and (b) "A p o i n t lies on a line" and "a line c o n t a i n s a p o i n t " a r e two forms of saying the same thing. a . The d e f i n i t i o n s : ~ ?collinear and coplanar are phrased using the second form. Rephrase these d e f i n i t i o n s using the f i r s t form. b. The f i r s t part of P o s t u l a t e 5 I s phrased u s i n g the second form. Rephrase t h i s part of Postulate 5 using t h e F i r s t form As in Problem 6, Postulate 1 Is written in one of the two forms Which form? Restate Postulate 1 in the other form. b.
5,
.
*6.
.
*7.
By P o s t u l a t e 5 a plane contains a t least three p o i n t s . Does it contain any more? On the b a s i s of our present postulates we cannot conclude t h a t it does, so we introduce
Postulate 6 . If two points l i e in a plane, then t h e l i n e c o n t a i n i n g these p o i n t s l i e s in t h e same plane. This postulate essentially says that a plane is flat, that is, that if it contains part of a line it contains the whole line. Theorem 3-2. If a line Intersects a plane not c o n t a i n i n g it, then t h e Intersection i s a single point. This follows from Postulate 6 i n the same way that Theorem 3-1 follows from P o s t u l a t e 1.
The figure shows line L, intersecting a plane
E in a point P.
You are going to see lots of drawings like t h i s , of figures in space, and to learn to draw them yourself. You should examine them carefully t o see how they work. We usually Indicate a plane E by d r a w i n g a rectangle in E. Seen L n perspective, the rectangle looks somewhat like a parallelogram. The l i n e L punctures E at I?. Part of L I s dotted. This I s the part t h a t you "can't seett, because the rectangular piece of E gets I n the way. (For a discussion on drawing 3-dimensional figures see Appendix V . ) We have seen that two points determine a l i n e . The next postulate s p e c i f i e s a similar determination of a plane.
1 1
Postulate 7 . Any three points l i e in at least one plane, and any three nun-collinear points l i e in exactly one plane. More b r i e f l y , any three p o i n t s a r e coplanar, and any three non-colllnear points determine a plane.
1
Theorem 3-3. Given a l i n e and a point not on the line, there is exactly one plane containing both of them.
The figure shows a plane E determined by the line L and the
p o i n t P. Theorem 3-4. Given two intersecting lines, t h e r e is exactly one plane containiry them.
P.
The figure shows two lines L, and Lp, intersecting in a point E I s the plane that contains both lines. Finally, w e state one more postulate;
i
P o s t u l a t e 8. If t w o d i f f e r e n t planes i n t e r s e c t , then their inter~ectlonis a lme.
Problem S e t 3-lc 1.
How many planes can contain one given point? t h r e e given points?
two given points?
On a level floor, why will a four-legged table sometimes rock, while a three-legged table is always steady? Complete; Two different l i n e s may intersect in a 9 and two d i f f e r e n t planes may Intersect in a Can t w o p o i n t s be non-colllnear? three points? f o u r points? n points? Write a careful definition of a set of non-colllnear points. Given: 1. P o i n t s A, 3, C lie in plane E. 2. Points A, B, C lie i n p l a n e F. Can you conclude that plane E is the same as plane F? Explain. Complete the following statements using the accompanying diagram. a. Points D, C , and are colllnear. b . Points E, F, and - are col linear, c. Points B, -, and A are collinear d. P o i n t s A , B , C , D , E , F , a r e A E
-
.
AB
Examine the following figure of a rectangular s o l i d until you see how it looks as a three-dimensional drawing. Then close the book and draw a figure l i k e this for yourself. Practice until you a r e s a t i s f i e d w i t h the results.
After doing Problem 8, draw a figure that represents a cube. Draw a plane E, using a pa-rallelogram to indicate t h e plane.
Draw a l i n e segment which lies in the plane E. Draw a l i n e that Intersects the plane E but does not intersect the line
11.
12.
seg-
ment. Use clashes to represent the part of the line hidden by the plane. The accompanying figure is a triangular pyramid, or tetrahedron. It has four vertices: A, B, C, D, no three o f
which are colllnear. a. Make a d e f i n i t i o n of an edge of t h i s tetrahedron. Use the I d e a s of t h e t e x t to h e l p you form the d e f i n i t i o n . b. How many edges does the tetrahedron El have? Name them. c . Are there any pairs of edges t h a t do not Intersect? d . A face is the triangular surface determined by any three vertices. There are four faces; ABC, AED, ACD, BCD, Are there any pairs of faces that do not Intersect? Explain. How many different planes (determined by D t r i p l e t s of labeled p o i n t s ) are there in the pyramid shown? Make a complete list, (YOU should have seven planes. ) ^"^
D> #
^^
A 8 3-2. Theorems in the Form of Hypothesis and Conclusion. Nearly every theorem is a statement t h a t If a certain thing is true, then something else is also true. For example, Theorem 3-1 states that if Ll and L2 are two different lines, then Ll Intersects Lp in at most one p o i n t . The if part of a theorem i s c a l l e d the hypothesis, OF t h e ~ 5 v e ndata, and the then part 1s called the Eelusion, or the thing to be proved. Thus we can write Theorem 3-1 in this way: Theorem 3-1. Hypothesis: Li and L2 are two different lines. Conclusion: L., intersects L2 in at most one point. Postulates, of course, are like theorems, except t h a t they are
----
-
--
-
-
.
not going to be proved. Most of them can be put in the same if then form as theorems. Postulate 1 can be stated t h i s way: Hypothesis; F and Q are two different p o i n t s . Conclusion: There is exactly one line containing P and Q. There a r e cases in which the hypothesis-conclusion form does n o t seem natural or useful. For example, the second part of Postu-
-
late 5, expressed in this form, looks awkward: Hypothesis: S is space. Conclusion: Not a l l points of S are coplanar. Such c a s e s , however, are very rare. It is not necessary, of c o u r s e , that all theorems be stated I n the hypothesis-conclusion form. It ought to be clear, regardless o f the form in which t h e theorem is s t a t e d , what part of it is t h e hypothesis and what part is the conclusion. It is very important, however, that we be able to s t a t e a theorem i n t h i s form if we want t o , because if we cannot, t h e chances are t h a t we do not understand exactly what the theorem says.
Problem Set 3-2
Indicate which part of each of the following statements is the hypothesis and which part is the conclusion. If necessary, rewrite in if-then form f i r s t . a . If John is ill, he should see a d o c t o r . b. A person w i t h red hair is nice to know. c. Four points are collinear if they lie on one line. d. If I do m y homework well, I will get a good grade. e . If a s e t of p o i n t s lies inone plane, the points are coplanar. f. Two intersecting lines determine a plane. 2. Write t h e following statements in conditional, or if-then, form: a . Two different l i n e s have at most one point In common. b. Every geometry student knows how t o add integers. c. When it rains, it pours. d . A l i n e and a point not on the l i n e are contained in exactly one plane.
1.
unethical. f. Two parallel lines determine a plane. Using the words "lf" and "then", write Postulate 1 and Theorem 3-1 in conditional form. Indicate the hypothesis and the conclusion f o r each case. a. Does the following statement mean the samething a s Theorem 3-4? " ~ w o lines always intersect in a point, and there I s exactly one plane containing them." Why or why not? b. Write Theorem 3-4 in the "hypothesis and conclusion" form. e.
3.
4.
3-3.
A dishonest practice is
Convex Sets.
Definition. A set A is called convex if f o r every two points P and Q of A, the entire segment l i e s In A . For example, the three sets pictured below are convex.
Here each of the sets A, B and C consists of a region of the plane. We have Illustrated their convexity by showing a f e w segments F. None of t h e s e t s D, E and F below is convex:
D
E
F
We have shown why not, by showing pairs of points P, Q for which t h e segment ^5 does not lie entirely in the g i v e n set. A convex set may be very large. For example, take a line L in a plane E and let H, and H2 be the sets lying on t h e t w o sides of L, l i k e t h i s :
The t w o sets Hi and H p are called half-planes or sides of L, and t h e l i n e L is called an edge of each of them. (Notice that L does not l i e in either of the two half-planes; L is not on either s i d e of Itself . ) If two points P and Q are in the same half-plane, say HI, then the segment 7S[ a l s o lies in Hi, and so does not intersect L.
Thus Hz is convex. And in the same way, Hg is convex; t h i s is illustrated by the points R and S in the figure. We notice, however, that if T and U are points belonging to different half-planes, then the segment intersects L, because you cannot g e t from one side of L to the o t h e r side without crossing the edge. We express t h i s fact by saying that I. separates H, from H2 in the plane, o r that L separates the plane i n t o t w o half-planes
Hl and Hz.
T h i s discussion I s a fair account of t h e facts, but it is not very good mathematical form, because It is based on a postulate that w e haven't even stated so far. We shall therefore state t h e p o s t u l a t e t h a t is needed, and then s t a t e the definitions that are based on it.
Postulate 9 . (The Plane Separation Postulate. ) Given a line and a plane containing I t . The points of t h e plane that do not l i e on the l i n e form two s e t s such t h a t (1) each of t h e s e t s is convex and (2) if P is in one s e t and Q is In the o t h e r then t h e segment T^ i n t e r s e c t s t h e line.
Definitions. Given a line L and a plane E containing It, the two sets determined by Postulate 9 a r e called half-planes, and L I s called an edge of each of them. We say that L separates I3 I n t o the two half-planes. If two points P and Q of E l i e In the same half-plane, we say t h a t they lie on the same side of L; if P lies In one of t h e half-planes and Q in the other they l i e on opposite sides of L. We see that the Plane Separation Postulate says two things about the two half-planes into which a l i n e separates a plane: (1) If two p o i n t s l i e in t h e same half-plane, then the segment between them lies in the same half-plane, and so never Inters e c t s the line. (2) If two p o i n t s l i e in d i f f e r e n t half-planes, then the segment between them always Intersects the line. If we do not restrict o u r a t t e n t i o n t o a single plane we can have many half-planes with t h e same edge. The picture
----
-
-
f i v e o f the infinitely many p o s s i b l e half-planes having line L for edge. Note that p o i n t s P and Q, although they l i e in d i f f e r e n t half-planes, cannot be said to be on o p p o s i t e sides of L. This c a n only be said of points l i k e P and R which are coplanar with L.
A plane separates space, in exactly the same way, i n t o two convex s e t s c a l l e d half-spaces. P
In the figure, Hi is the half-space above E and H p i s t h e halfspace below E. P and Q l i e in H,, and so a l s o does the segment 7^. P and S are in different half-spaces, so t h a t the segment 7 S ' i n t e r sects E in a point X. R and S are in the same half-space H2, and so also I s the segment E.
This situation is described in the following postulate.
.
Postulate 10. (The Space Separation Postulate ) The p o i n t s of space t h a t do not lie In a given plane form two s e t s such t h a t (1) each of the sets I s convex and ( 2 ) if P is In one set and Q is in t h e other, then t h e segment P^ intersects t h e plane.
Definitions. The two sets determined by Postulate 10 are called half-spaces, and t h e given plane I s called the face of each of them. Note that while a line is an edge of i n f i n i t e l y many h a l f p l a n e s , a plane is a face of c n l y two half-spaces,
-
-
Problem S e t 3-3
In answering the following q u e s t i o n s use your intuitional understanding of planes and space in situations not covered by our p o s t u l a t i o n a l structure. 1. Be prepared to discuss the following questions orally. Is a l i n e a convex set? Explain. Is a s e t consisting of only two p o i n t s convex? Why? Is a ray a convex set? If one point I s removed from a line, do the remaining p o i n t s form a convex s e t ? Why? Is the s e t of p o i n t s on the surface of a sphere convex? Why?
Is the space enclosed by a sphere a convex s e t ? Does a point separate a plane? space? a line? Does a r a y separate a plane? Does a line? Does a line segment? Can two lines in a plane separate the plane i n t o two regions? Three regions? Four regions? F i v e regions? I n t o how many parts does a plane separate space? What are these parts called?
Every point on ?Q is contained In t h e s e t shown. Does t h i s mean t h a t t h e s e t I s convex? Explain. Which of the regions indicated by Roman numerals are convex s e t s ? Give reasons f o r your choice.
Is every plane a convex s e t ? Explain. Which postulate I s essential in your explanation? The interiors of c i r c l e s A and B are each convex sets. a. Is t h e i r Int&*secfcion a convex set? Illustrate b. Is their union a convex set? Illustrate. If one point is removed from a plane, is the s e t formed convex? Why? If L I s a l i n e in a plane E, i s the set of a l l points of E on one side of L a convex set? Draw a plane quadrilateral ( a figure w i t h f o u r sides ) whose i n t e r i o r 1s convex. D r a w one whose i n t e r i o r I s not convex. Is the s e t of points containing all points on the surface and a l l paints i n ?!be i n t e r i o r of a sphere convex? Is the set of p o i n t s in a t o r u s ( a doughnut shaped figure) convex? Is the union of t w o half -planes which a r e c o n t a i n e d in a plane the whole plane if a. the half-planes have the same edge? Explain. b. the edge of one half-plane intersects t h e edge o f the other half-plane In exactly one point? Explain, using a diagram if necessary. a. Into how many parts does a point on a line separate t h e line? What name would you suggest giving to each of these
.
b.
'3D
parts? Using t h e terminology you developed In part ( a ) , write out a Line Separation Statement similar to Postulates9 and 10.
13. How does a ray d i f f e r from a half-line? 14. Can three lines in a plane ever separate the plane into three regions? f o u r regions? f i v e regions? s i x regions? seven regions? 15. Into how many parts do two intersecting planes separate space? Two parallel planes? 16. What is the greatest number of parts i n t o which space can be separated by three distinct planes? What I s the least number? *17. Wri-te a careful explanation of why the following statement is true. The intersection of any two convex sets which have at least two p o i n t s i n common is convex. (Hint: L e t P and Q be any two points belonging to the intersection.) *l8. Sketch any geometrical solid bounded by plane surfaces such that t h e s e t of points in the interior of the figure is not convex.
Review Problems 1.
2.
3.
4.
Each of 3 planes intersects each of the others. May they intersect i n one line? Must all three intersect in one l i n e ?
Explain. How many planes will contain the three given points A, B, and C if no l i n e contains t h e m ? Write each of the following statements In the "if -then" form. a. Zebras with polka dots a r e dangerous. b. Rectangles whose sides have equal lengths are squares. c. There will be a c e l e b r a t i o n if Oklahoma w i n s . d . A plane is determined by any two intersecting lines. e. Cocker spaniel dogs a r e sweet tempered. Supply the following information about the postulates in the chapter. What p r o p e r t y of each of the half-planes i s mentioned in the Plane Separation Postulate? Do the half-spaces of the Space Separation P o s t u l a t e have the same property?
C r i t i c i z e the following statement: 'The t o p of the table is a plane." List a l l the situations we have studied which determine a sing l e plane. A s e t is convex If f o r every p a i r of points in It, all p o i n t s of t h e segment Joining the two p o i n t s l i e Given that plane E separates space into half-spaces R and t h a t point A is in R and point B is in S, does Intersect E? L, intersects plane E in P but does not l i e In E. LA plane E but does not contain point P . Is I t possible and Lp to intersect? Explain. a . A s e t o f points is colllnear if b.
and S, have to lies.in for L,
A set of points i s coplanar i f
.
May 5 p o i n t s be c o l l l n e a r ? d . Must 2 p o i n t s be c o l l l n e a r ? e . May n points be colllnear? f Must 5 points be coplanar? g . May n points be coplanar? Points P and Q l i e In both planes E amd F which i n t e r s e c t d n # line AB. Would It be correct to say that P and Q l i e on AB? c
.
Explain. Is the union of a half-plane and a ray on its edge convex?
Chapter 4 ANGLES AND TRIANGLES
4-1.
The Basic Definitions. -An angle is a figure that looks l i k e one of these:
To be m o r e exact: An angle 1s the unlon of two mys whlch have t h e same end-point but do not l i e in the same line. The two ray3 are called t h e sides of the angle, and their common endpoint I s called the vertex. + + The angle which is the union of A 3 and AC is denoted by Definitions,
LMC, or by LCAB, or aimply are meant. Notice means of A and any
by
that/^^^ two
A
LA If I*
l a clear which rays
can be equally w e l l d e s c r i b e d by points on different s i d e s of t h e angle.
C
E
In the above f i g u r e LDAE l a the same as /_BAC, same as Afi and 13 the same as 3.
a
+
because AD is the
Notice that an angle goes out Infinitely far In two direct l o n s , because I t s sides are rays, rather than segments. The f i g u r e on the left, below, determines an angle uniquely, but I s not all of the angle; to get a l l of the angle, we have to extend -> ->Â and AC g e t t i n g rays AB and AC, as on t h e r i g h t . the segments
Definitions. If A, B, and C are any three non-collinear and AC I s called points, then the union of the segments
a
a triangle,
and is denoted by AABC; the points A, B and C are called I t s vertices, and the segments AB, BC and AC are called Its sides. Every triangle determines three angles; AABC determines the angles LING, LABC and whlch are called the angles of AABC. For s h o r t , we w i l l often write them simply a a /A, /B, and Note that while AABC determines these t h r e e angles, I t does not actually contain them. J u s t as a school does n o t c o n t a i n I t s own graduates, so a triangle does n o t contain Its own angles, because t h e s i d e s of a t r i a n g l e are segments, and the sides of an angle are r a y s . To draw the angles of a t r i a n g l e , we would h a v e t o extend the sides of the triangle t o g e t rays, like t h i s :
-
--
-
LACB,
LC.
There usually Is not much point in doing t h i s , however, because it is plain what the angles of a triangle are supposed to be.
[sec. 4-11
The interior of an angle consists of all points that l i e Inside the angle; and the exterior of an angle consists of all the points t h a t l i e outside, l i k e t h i s :
/
Exterior
-
#'
Exterior
Interior
*s
We can s t a t e t h i s more exactly as follows: Definitions. Let /BAC be an angle lying in plane E. A point P of E lies In the interior of /BAC If (1) P and B are on t h e same s i d e of the line and a l s o (2) P and C are 0 is the on the same s i d e of the line AB. The e x t e r i o r of /BAG s e t of all points of E that do not l i e I n the I n t e r i o r and do n o t l i e on the angle i t s e l f . You should check carefully t o make sure that t h i s really says w h a t we want it to say. In the figure, P l a In the i n t e r i o r , ¥ because P and B are on the same s i d e of AC and also P and C are on the same side of %. Q is in the exterior, because Q and C <¥ are not on the aame a i d e of A3. R l a In the exterior, because <^-fr> R l a on the "wrong side" of both o f the lines A 3 and AC. S is in 0 the exterior because I t is on t h e "wrong s i d e H of AC. Notice that we have d e f i n e d the interior of an angle as the intersection of two half-planes. The half-planes look l i k e t h i s :
-
Here one of t h e half-planes is cross-hatched horizontally, the other I s cross-hatched vertically, and the Interior of /BAc is cross-hatched both ways. The interior of a triangle c o n s i s t s of the points t h a t lie i n s i d e it, like this:
More precisely: Definitions. A point lies In the i n t e r i o r of a triangle if It l i e s i n the I n t e r i o r of each of t h e angles of t h e triangle. A point lies I n the exterior of a triangle If I t lies in the plane of the triangle but Is not a point of t h e triangle or of I t s Interior. You should check c a r e f u l l y to make sure t h a t t h i s r e a l l y says w h a t we want it to say.
--
Problem Set 4-1 Complete this definition of angle: An angle I s the which have the same endof t w o point, but do n o t l i e in t h e same Complete this definition of triangle: A triangle I s the of the three joining each p a i r of three points C Are the sides AC and of AABC the same as the sides of LA? Explain,
.
.
A B Is t h e union of two of t h e angles of a t r i a n g l e t h e same as the triangle I t s e l f ? Why? Into how many regions d o the angles of a triangle separate the plane of the triangle? ',M Complete:
Name the angles In the figure.
HOW many angles are determined
by the figure? Name them. How many may be named using the vertex l e t t e r only? A
8
9. Name the angles in the figure.
here
are more than s i x . )
10. Name a l l the triangles in the figure.(There are more than eight).
11.
Name the points of the *D figure which are in the IneF terlor of /CBA. OM b. Name the points of the N figure In the exterior B A G Of /B. 'H Is the vertex of an angle in the Interior of t h e angle? in the exterior? Explain. Is the Interior of an angle a convex s e t ? l a the e x t e r i o r ? Is a t r i a n g l e a convex s e t ? I s the interior o f a t r i a n g l e a convex s e t ? is t h e exterior? a. Can a point be in the exterior of a triangle and In the i n t e r i o r of an angle of t h e triangle? Illustrate. b. Can a point be In t h e exterior of a t r i a n g l e and not In the i n t e r i o r o f any angle of t h e triangle? Illustrate. Given AAEC, and a p o i n t P. P is in t h e i n t e r i o r of LBAC and a l s o in the I n t e r i o r of /ACB. What can you conclude a b o u t point P? GivenAABC and a p o i n t P. P and A are on the same side of 8. P and 3 are on the same a i d e of E. a.
-
12. 13.
14. 15. 16.
l7+
18.
-
the i n t e r i o r of LACB? b. Is P in the interior of AABC? 19. Carefully explain w h y the following statement 18 true: If a line m Intersects t w o s i d e s C of a t r i a n g l e ABC in points D and E, not the vertices of the triangle, then line rn does not i n t e r s e c t the third side. Show t h a t A and B are (Hint: I n t h e same half-plane. )
a*
1 s P in
-
Remarks On Angles. What we have presented in t h i s chapter is the simplest form of the idea of an angle. According t o our d e f i n i t i o n , a n angle i s simply a s e t which is t h e union o f two non-collinear my&
4-2.
Angles, I n t h i s aense, w i l l be quite good enough for the purposes of t h i s c o u r s e . Later, you w i l l see the Idea of an a n g l e I n vari o u s o t h e r forms. Here we explain these other forms b r i e f l y , merely I n order to a v o i d confusion In case you may have heard o f them already. (1) In the f i r s t place, we sometimes t h i n k of an angle as being o b t a i n e d by rotating a ray from one position to a n o t h e r . In t h i s case, one r a y i s the i n i t i a l s i d e , and the o t h e r is t h e tenalnal side. Thus we would c o n s i d e r the two angles below as being d i f f e r e n t , because t h e r o t a t i o n s are i n two d l f ferent
The first Is called a p o s i t i v e angle; the rotation l a counterclockwise. The second I s a n e g a t i v e angle; the r o t a t i o n la clockwise.
(2) People sometimes speak of s t r a i g h t angles, which look l i k e this:
Here the rays A^ and 8 are considered t o form an angle, even though A, B, and C are collinear. (3) F i n a l l y , we sometimes distinguish between an ordinary angle and a reflex angle having the same rays as its sides. The double-headed arrow below is supposed t o I n d i c a t e a reflex angle;
These complications, and various others of the same s o r t , w i l l not be used in t h i s book, because they w i l l not be needed. F o r example, t h e angles o f a triangle are never reflex angles, and there I s no reasonable way to d e c i d e in which direction they should be considered to go. Not until we g e t t o trigonometry do these fancy angles become necessary and Important
.
-
Measurement Of' Angles. Angles are usually measured In degrees, with a p r o t r a c t o r . With t h e protractor placed as in t h e f i g u r e below, with its edge on the edge of the half-plane H, we can read o f f the measures of a large number of angles. 4-3.
Figure A. The number of degrees In an angle l a c a l l e d i t 8 measure. there are r degrees In t h e angle /XAY, then we write m / XAY = r. For example, In the figure we read off t h a t m /PAB = 10,
-
m / W = 40, m / RAB = 75, m LSAB = 90, m / TAB = 105, Of course, the rays t h a t are drawn form more angles By subtraction, we can see t h a t m /W = 40 10 = 30, m /SAR = 90 75 = 15,
-
and so on. than t h i s .
-
-
and so on.
If
-
Since m / QAB = 40, we speak of /QAB as a we i n d i c a t e Its measure in a figure l i k e this:
angle, and
~ 1 0
need t o uae the degree sign when we write ~ / Q A B = 40, because as we explained at the outset, rn /QAB means the number of degrees in the angle. Notice t h a t in Figure A there is no such t h i n g a s the angle /CAB, because the r a y s A? and Afi are collinear. But we notice -> that the ray AC corresponds to the number 180 on the number-scale of the p r o t r a c t o r , and the ray 3 corresponds to t h e number 0. Therefore we can find m LCAU by w r i t i n g m /CAU 180 130,
But we don't
-
= 50.
Similarly, ITI
LCAQ = 180 -
40,
= 130.
The following postulates merely summarize t h e facts about p r o t r a c t o r s that we have Just been discussing. Each of them I s I l l u s t r a t e d by a f i g u r e .
P o s t u l a t e 11. To every angle /BAG between 0 a n d 180.
h he
Angle Measurement P o s t u l a t e . ) t h e r e corresponds a real number
Definition. The number specified by P o s t u l a t e 11 is c a l l e d t h e measure of the awle, and written as m L B A C .
--
f
1 1 1
Let
A?
P o s t u l a t e 12. (The Angle Construction P o s t u l a t e . ) be a ray on the edge of the half-plane H. For
every number r between 0 and 180 there is exactly one ray A?, with P In H, such t h a t m /PAB = r.
(The Angle Addition p o s t u l a t e . ) If D is a point in the i n t e r i o r of D A C , t h e n m@G = m@D + ~/_DAc.
Postulate 13.
on this basis t h a t we computed the measures of angles by s u b t r a c t i o n , w i t h a p r o t r a c t o r placed w i t h I t s edge on t h e -> = ~ ~ B -A CBAD.) ray AB. (~LDAC Two angles form a l i n e a r pair i f t h e y l o o k l i k e this:
In
was
That is : -> Definition. If A^ and A^ are opposite r a y s , and AD Is a n o t h e r ray, then /-BAD and ~ A form C a linear pair. Definition. If t h e sum of t h e measures of two angles is 180, then t h e angles are called supplementary, and each is c a l l e d a supplement of the o t h e r . Hence the name of t h e following p o s t u l a t e *
P o s t u l a t e 14, h he Supplement P o s t u l a t e . ) If two angles fo m a linear pair, then they a r e
1
supplementary.
--
Problem Set 4-3
\
1.
Using t h e figure, find <,he value of each of t h e following: a, m L F A B . g. m L E A D . h. FAG + KI/_GAH. b. m L E A B . 1. RI /GAF + m /FAE. C. m /MAC. j. m LMAB m LFAE d. ~/FAE. k . rn/-mB m/-DAB. e . m LGAE. 1. m LMAB n~LNAH. f. MAN. 2. With continued p r a c t i c e you should "be able t o estimate t h e s i z e of angles f a i r l y a c c u r a t e l y without using a protractor. Do not use a p r o t r a c t o r to d e c i d e which of the angles shown have measures within t h e indicated ranges.
-
-
Match the corresponding pairs :
3.
4.
5.
-
Using only a straightedge and not a p r o t r a c t o r , sketch angles
whose measures are approxlJnately 30, 150, 45, 60, 135, 90, Then use your p r o t r a c t o r to check your sketches. On the edge of a half-plane, take a segment about 3 Inches long. A t A draw ray A?S In the half-plane f o r m i n g / ~ ~of~ A t B draw ray B& in the same half-plane forming /ABD 58'. of 72O. Measure the remaining angle of the triangle fonned.
In the figure, a. m / _ B H F + m / J 3 V P a g i / _ to. r n / G F H + m / . B P H = r n ^ _
? ? 8
6.
In the figure, a. ~/XZK + ~ ^ _ K z R + IR/_YZR a m/_ b. m/XZR m / R Z K = m/ ? m/ XZY rn/XzK ss m i ? c. d. If Y, R, K and X are c o l l i n e a r , then m/ YRZ + rn^JSRx = ?
- -
-
-
?
and C6 Intersect forming four angles. Using t h e Indicated measure, find a, b and c .
In the figure,
Determine the supplement of each of',the following: lloO, goo, 36') 15.5', no, (180 n)', (90 n)¡ If one of two supplementary angles has a measure 30 more t h a n t h e measure of t h e o t h e r , what is the measure of each angle? If the measure of an angle is twice t h e measure of its supplement, f i n d the measure of t h e angle. The measure of an angle is f o u r times t h e measure o f i t s supplement. Find t h e measure of each angle. -> a. Given a ray AC lying on the edge of a half-plane H, and a number r between 0 and 180. In how many ways can you -> construct a ray AB In H such that r n 1 3 A = ~ r? Why? -> b. Given a ray AC l y i n g in a plane EJ and a number r between 0 and 180. In how many ways can you construct a ray in E such that ~/BAC = r? Why?
-
-
3
4-4.
Perpendicularit& R i a t Angles
and Con~ruence of -
Angles.
Definitions. If the two angles of a l i n e a r p a i r have the aame measureJ then each of the angles is a right angle.
Since r + r = 180, by the Supplement Postulate, we see t h a t Th33 can be regarded as an a ri~$~t angle is an a n ~ l eof alternative definition of a right angle; It i s equivalent t o our
--
-
&.
first d e f i n i t i o n . In terms of right angles it is easy to define perpendicularity of any combination of line, ray or segment. In applylng the following definition remember t h a t a ray or a segment determines a unique line which contains i t . Definition. Two intersecting s e t s , each of which i s either a l i n e , a ray or a segment, are perpendicular i f the two l i n e s which contain them determine a right angle. Definition. If t h e sum of the measures of two angles is 90, then the angles are called complementary, and each of them i s called a complement of the other. (Compare t h i s with the d e f l n l t i o n of supplementary amleg, J u s t before the statement of the Supplement Postulate ) An angle with measure leas than 90 I s called -' a c u t e and an angle with measure greater than 90 i s called obtuse.
.
Obtuse
,I
/
Definition. Angles w i t h t h e same measure are called congruent angles. T h a t Zs, / BAC and are congment if = ~LFQR. In t h i s case we write
-
LPQR / BAC -
~LBAC
2 /PQR.
Notice t h a t the e q u a t i o n m/ -BAG = ~ / P Q R and t h e congruence / BAC 2 / PQR are completely equivalent; we can replace one by L the other any time we want to.
-
The following theorems are easy to prove, if we remember
clearly what the words mean: Theorem 4-1. If t w o angles are complementary, then both of them are a c u t e . Theorem 4-2. Every angle is congruent to i t s e l f . Theorem 4-3. Any two right angles a r e congruent. Theorem 4-4. If t w o angles are both congruent and supplementary, then each of them is a right angle. ( ~ i n t : L e t r be the number which is the measure of each of the t w o angles, and then f i n d out what r must be. ) Theorem 4-5. Supplements of congruent angles are congruent. Restatement: If (1) 2 ( 2 ) LA and L 3 are ~upplementary and [ 3) and are supplementary, then
LB LD, L C LD
?roof: The statement t h a t L Ei 2 L D means that m L B and and m/ D are t h e same number r, as in the figure. Since LA L 3 are supplementary, it follows t h a t
-
For t h e same reason, m/C
rnL~
= 180
-
m/D
= 180
- r.
LA LC.
Therefore = m k C , which means that 2 You must not conclude from t h e above picture that supplementary angles must necessarily be placed beside one another* In a way that makes it evident t h a t their measures add up to 180.
[ s e e . 4-41
The following p i c t u r e a l s o serves to illustrate Theorem 4-5.
In drawing pictures to Illustrate theorems or problems you should r e a l i z e that the figure in the book is not t h e only correct one, and you should try t o make your p i c t u r e d i f f e r e n t from t h e one given In t h e book. Theorem G . complement^ of congruent angle3 are congruent. The proof of t h e theorem I s exactly analogous to the preceding proof, and you should write I t out for yourself. When two lines Intersect, they form four angles, l i k e t h i s :
L2
are c a l l e d v e ~ t l - 1 angles, and and L 4 are a l s o L l and c a l l e d vertical angles. More p r e c i s e l y : D e f i n i t i o n . Two angles are vertical angle^ if their sides form two pairs of opposite r a y s . It looks as if these pairs of vertical angles ought to be congruent, and I n fact t h i a l a w h a t always happens : Theorem 4-7. V e r t i c a l angles are congruent.
3 are
opposite rays, and & and A^i are opposlte ray3, so m a t L l and L 2 are vertical angles. Then L l and L 3 are supplementary, and L 2 and /,-3 are supplementary.
Proof:
Given t h a t
and
and L 2 Since L 3 is congruent to itself, t h i s means that l 2 L 2 , whlch have congruent supplements. By Theorem 4-5, L was to be proved. If two intersecting lines form one right Theorem 4-8. angle, then they form f o u r r i g h t angles.
-
t
You should be able to supply t h e proof.
Problem Set 4-4 1.
2.
In a plane, how many perpendiculars can be drawn to a line at a given point on t h e line? b. In space, how many perpendiculars can be d r a w n to a line at a given point on the l i n e ? +' 4 If OR and OS are opposite rays and 6% is a r a y such t h a t <-Ã m /RON = m /SON, what can you conclude about 8k and RS? Explain. In half-plane H, % and 3 a r e opposite rays, m L R X B = 35 and a.
= 90. Name a pair o f perpendicular rays, If any occur in t h e figure Name a pair of complementary angles, if any occur in the figure. Name a pair of v e r t i c a l angles, If any occur in t h e figure. Name two pairs of supplementary angles In the figure.
m /RXS
a.
.
b.
c. d
.
Determine the measure of a complementary angle f o r each of the following: 0 a. .'01 d. x * b. c.
a.
b. a
8 0 ' .
-
(90 X I ' . f. (180 ,')x 44.5'. If two angles w i t h t h e same measure are supplementary, e.
-
what is the measure of each? If two angles w i t h t h e aame measure are complementary, what is the measure of each? If two lines intersect, how many pairs of v e r t i c a l angles
are formed? b. If t h e measure of any one of the angles in (a) is 70, what is the measure of e&ch of t h e others? If all of the angles In ( a ) are congruent, what i s t h e c. measure of each? If one of a pair of vertical angles has a measure of r, write t h e formulas f o r t h e measures of the other three angles formed In half-plane H, & and & a r e opposite r a y s ,
.
Prove Theorem 4-1. A G P r o v e Theorem 4-4, Given: In the figure f o r Problem 8, & -L are opposite rays and /DOE are complementary. Prove; /AGB
.
E
tf& and Ok
and
&
In plane E, lines 8, %, I n t e r s e c t at 0.
Given:
S,
Prove:
13.
14. 15.
%
b
+
g
+
& and
d = a.
-Ã
OC are three different r a y s in a plane, no two of them opposite, indicate true or f a l s e f o r each of the f o l l o w i n g statements and explain your answer. a. ~ / _ A O B + m/JBOC = ~ / _ A o c . + m LBOC + rn /AOC = 360. b. rn LAOB The measure of an angle is n i n e times t h a t of I t s supplement. What is the measure of the angle? A l a y o u t drawing l a a plane drawing which can be folded to form t h e boundary of a given solid. Below Is pictured a cube and a layout drawing for i t .
If
and
- -u
.
( ~ o t t e dlines I n d i c a t e f old3 ) Use your Imagination, your ruler and your p r o t r a c t o r t o make a l a y o u t drawing f o r each o f the figures below. Then cut o u t your drawing, f o l d on d o t t e d l i n e s , and t a p e t o gether. Use c a r d b o a r d o r heavy paper f o r a rigid figure. a , A pyramid whose base is 8 square w i t h 2" sides and whose o t h e r faces are isosceles triangles w i t h 60Â base angles. 2"
( Problem 15 continued) b. A prism whose bases are pentagons w i t h 1 i n c h s i d e s and 108' angles, and whose height I s 2 Inches.
1. 2.
3. 4.
5.
6. 7. 8.
9. 10.
11. 12.
13.
Review Problems What tool is used to measure angles? To every angle there corresponds a real number between and , called the measure of t h e angle. An angle w i t h a measure of less than 90 is Two angles formed by the union of two opposite rays and a t h i r d ray a l l w i t h the same end point are a of angles. If the sum of the measure of two angles is 90, then each is called a of the other. A n angle with a measure greater t h a n 90 is called Angles with t h e same measure are If two angles are b o t h congruent and supplementary,then each of them is a Supplements of congruent angles are If t w o angles are complementary, then each of them Is
An angle is t h e of two which have a common end point. If X, Y, Z are t h r e e points, t h e union of the t h r e e segments connecting them In pairs l a a A point X is In the interior* of RST if points R and lie on the same s i d e of and if points X and l i e on the same side of If the sum of t h e measures of two angles l a they are
.
14-
c a l l e d complementary and If t h e sum is
they a r e
called
15.
16,
Two opposite angles formed by two Intersecting lines a r e angles. They are always congruent. andm opposite rays. The points E, F, and H are on t h e same side of AS* P o i n t s E and H are on opposite aides of %. Points A and H are on t h e same side of 1 8 1 % and B I F B E = 20. D r a w t h e figure and f i n d : a. rn /EBA. b. m /FBH. c. m /EX.
AS
.
313.
8
-
17.
Given: m /BCD III
/BOC
m /DCO m /DAO
Find : = 90, = 50, = 25, = 45.
a. C.
m /-DOC. m /KO. m LDOA.
d.
~LAOB.
b.
If one of two supplementary angles has a measure of 50 more than the measure of t h e o t h e r , what is the measure of each angle? The measure of an angle is five times t h a t of i t s complement. Find the measure of each angle. Under what c o n d i t i o n s are the angles of a l i n e a r p a i r congruent ? Is there a p o i n t i n t h e p l a n e of a t r i a n g l e such that the point I s neither In t h e e x t e r i o r nor the I n t e r i o r of a triangle and n e i t h e r i n the i n t e r i o r nor the exterior of any of I t s angles? Is t h e measure of an angle added to the measure of an angle t h e measure of an angle? Explain.
23.
Could the interior of a triangle be considered as the intersection of three half planes? Illustrate.
c 24.
How many triangles are in t h i s figure?
25,
Does m
26.
Does
27,
Is
28.
How many angles are i n d i c a t e d in the drawing?
LBAC = m /^BAE? LBAC = /BAE?
LABE aupplernentary
to
LEBC?
Problems 24
29.
Explain c a r e f u l l y why the following If a line m i n t e r s e c t s 2 s i d e s of a, points U and V, not the vertices of line m does n o t i n t e r s e c t the t h i r d
<
31.
-
statement I s true: triangle ARST In the triangle, then side.
zAs
B C 1f' you were glven t h a t L a 2 / b and that / x 1s ~ ~ P P l e m e n t a ~ t o L a and that L y is supplementary to L b , what theorem or postulate would you use to prove that L x 2 / y? The Angle Measurement P o s t u l a t e places what limitation on angle measures? Is the following a correct restatement of the Angle Construction Postulate: Given a ray X? and a number k -> between 0 and 180 there i s exactly one ray XP such that m /FXY = k? Explain.
-
-
-
32,
33.
28.
34.
By giving its name, or by s t a t i n g I t in f u l l , give t h e postulate which seems to you to be most appropriate In each of the following cases, a s reason f o r the statement.
35.
Is the following statement always t r u e ? If !S and Intersect at 0, then /AOC 2 /BOD.
8
Chapter 5
C ONGRUENCES
5-1.
The Idea of a Congruence. -
Roughly speaking, two geometric figures a r e congruent if they have exactly the same s i z e and shape. F o r example, in t h e figure below, all three triangles are congruent.
One way of descrialnp; t h e situation is to say that any one of these triangles can be moved onto any other one, in such a way that it fits exactly. Thus, t o show what we mean by saying that two triangles are congruent, we have to explain what points are supposed t o go where. For example, t o move AABC onto A DFE, we should p u t A on E, B on F, and C on D. We can write down the pairs of corresponding vertices l i k e t h i s : A-E
B-F C-
D.
To describe the congruence of the first triangle and t h e t h i r d , we should match up t h e v e r t i c e s like t h i s ; AG 3-H . !-
c -I. How would you match up the v e r t i c e s to describe the congruence of the second triangle w i t h t h e t h i r d ? A matching-up scheme of thls kind is called a one-to-one correspondence between the vertices of the two triangles. If t h e matching-up scheme can be made to work - - t h a t I s , i f t h e
triangles can be made to f i t when the vertices are matched up in the prescribed way - - then the one-to-one correspondence is called a congruence between the t w o t r i a n g l e s . For example, t h e correspondences t h a t we have just given are congruences. On the other hand, if we write A-F
B-D c -E, this does give us a one-to-one correspondence, but does&n give us a congruence, because the f i r s t and second triangles cannot be made to coincide by this particular matching-up scheme. We can write down one-to-one correspondences more b r i e f l y , in one line. For example, the correspondence A-E B-I'
c -D, which i s t h e first example that we gave, can be w r i t t e n i n one l i n e l i k e this: ABC EFD.
-
Here it should be understood that t h e f i r s t l e t t e r on the l e f t corresponds to t h e first letter on the r i g h t , the second corresponds t o the second, and the t h i r d to t h e t h i r d , l i k e t h i s :
EFD
ABC
L e t us t a k e one more example.
These two f i g u r e s are of t h e same size and. shape.
To show how one can be moved onto t h e o t h e r , we s h o u l d match up t h e v e r t i c e s l i k e this: A-H
B-G
-
C-F
D
E.
These two f i g u r e s are congruent, because the correspondence t h a t we have w r i t t e n down is a congruence, t h a t is, t h e f i g u r e s can be made t o f i t If t h e v e r t i c e s are matched in the given way. For short, we can write the congruence in one l i n e , like this:
Notlce that t h e order in which the matching pairs are written does not matter. We could have w r i t t e n o u r list of matching pairs this way: D-E B-G C-
F
A-
H;
and we could have described our one-to-one correspondence In one l i n e , l i k e this: DBCA EGFH A l l that matters Is which p o i n t is matched w i t h which. It i s q u i t e p o s s i b l e f o r t w o figures t o b e congruent i n more than one way.
- .
Here t h e correspondence ABC
-
-
FDE
is a congruence, and the correspondence ABC
FED
is a d i f f e r e n t congruence between the same two figures . Obviously AABC c o i n c i d e s with itself. If we agree t o match every v e r t e x w i t h i t s e l f , w e g e t t h e congruence ABC
-
ABC.
This is called t h e i d e n t i t y congruence. There i s a n o t h e r way of matching up t h e vertices of t i l l s t r i a n g l e , however. We can use t h e correspondence ABC ACB. Under t h i s correspondence, t h e figure I s made to coincide with itself, with t h e v e r t i c e s 3 and C interchanged. This is not p o s s i b l e for all triangles by any means; it won't work unless at least two sides of t h e triangle are of the same l e n g t h .
-
--
Problem Set 5-1
I n t h e problems of t h i s s e c t i o n , t h e r e are no t r i c k s in t h e way that t h e figures a r e drawn. That is, correspondences t h a t l o o k l i k e congruences when t h e f i g u r e s are measured with reasonable care really are supposed to be congruences. In t h i s s e c t i o n we are n o t trying to prove things. We are merely trying to learn, informally, what the idea of a congruence Is all about. 1. Below there are s i x figures. Write down as many congruences as you can, between t h e s e figures. ( ~ not o count the ident i t y congruence between a figure and itself but recall that t h e r e is a congruence between a triangle having two congruent s i d e s and itself that is not the identity.) You should get 6 congruences in all. {One congruence is DEF -SUT.)
4.
Answer a s in Problem 1:
5. Name the f i g u r e s t h a t do n o t have a matching figure.
6.
Which p a i r s of t h e following figures are congruent?
7.
The t r i a n g l e below is equilateral.
That is, AB = AC = BC.
8.
For the triangle on t h e preceding page, write down all congruences between the triangle and i t s e l f , starting with t h e identity congruence ABC A B C , (you should get more than f o u r congruences.) Write down a l l of t h e congruences between a square and itself.
9.
a.
-
I f two f i g u r e s are each congruent t o a t h i r d , are they congruent to each o t h e r ?
Is a figure congruent to i t s e l f ? c. Can a t r i a n g l e be congruent t o a square? d . A r e t h e t o p and bottom faces of a cube congruent? e. Are two adjacent faces of a cube congruent? f Are t h e t o p and bottom f a c e s of a rectangular block, such as a brick, congruent? g. Are two adjacent faces of a brick congruent? 10. Pick out the pairs o f congruent figures, b.
.
106 1.1. Write down t h e four congruences of t h i s figure with itself.
12.
Suppose A, B, and C are three points of a l i n e as shown w i t h
AB = EC.
Describe a motion of the line that takes A t o where B was. Does it necessarily take B to C? b. Describe a motion of the line that interchanges A and C. Under what conditions c a n the following pairs of f i g u r e s be made to coincide by moving one in space without changing i t s size and shape? (it Is understood that this moving is done abstractly In the mind. One figure c a n move through another so t h a t a s o l i d can be moved onto another solid of the same s i z e and shape. For example, one segment can be moved to coincide w i t h another if they have the same length. One sphere can be moved to c o i n c i d e w i t h a n o t h e r if their radii are t h e same length. ) a. Two segments. b. Two angles. c. Two rays. d . Two c i r c l e s . e . Two cubes. f. Two p o i n t s . g. Two l i n e s . a.
14.
Given a c i r c l e containing t h r e e p o i n t s A, B, C as shown, w i t h t h e arc from A to B the same length as t h e arc from B to C.
a.
15.
Describe how the c i r c l e may be moved to take A to where B was and 3 t o where C was. b. Describe how t h e circle may be moved to leave B fixed but to Interchange A and C. Suppose t h a t the following ornamental f r i e z e extends infinitely In both directions, as a l i n e does.
a.
b.
Describe motions of two different t y p e s that Induce congruences of t h e frieze with I t s e l f . How many such congruences are there altogether? Do the same f o r this f r i e z e .
16.
Which of the following figures can be f i t t e d onto each other? For each matched p a i r , tell whether you must t u r n the f i g u r e over in space as well as s l i d e and rotate it in a plane to make it f i t on the o t h e r so that a l l segments f i t .
17. The f i g u r e below is a f i v e - p o i n t e d s t a r .
Write down a l l of the congruences between t h e star and itself. To save t i m e and paper, l e t us agree t h a t a congruence f o r this figure i s sufficiently described if we say where the p o i n t s A, B, C, D, E of the star a r e supposed to go. For example, one of t h e congruences that we a r e looking for can be written as ABODE -BCDEA.
5-2.
Congruences between Triangles. In the preceding s e c t i o n , we have explained the basic idea of what a congruence is. Let us now give some mathematical definitions so t h a t we can talk about congruence in a c a r e f u l way, in terms of distance and angular measure, instead of having to talk l o o s e l y about things falling on each other. For angles and segments, it is easy to say exactly what we mean : Definitions. Angles are congruent If they have t h e same measure. Segments are congruent If they have the same length. The first definition above is merely a repetition from Section 4-3. Analogous to Theorem 4-2 for angles we have a theorem f o r segments : Theorem 5-1. Every segment i s congruent to i t s e l f . We sometimes r e f e r to these two theorems by the term Identity congruence. Just as we i n d i c a t e t h a t L A and C B writing Z.A 2 LB, so we may write
are congruent, by
m2^S
AB
and CD are congruent. In t h e table below, t h e equation on t h e left and the congruence on the right in each l i n e may be used i n t e r c h a n g e a b l y : 1. m L A = m LB. 1. L A 2 LB. 2. AB = CD. 2. " S 5 ^ ^ 5 . Each of the equations on t h e left is an equation between numbers. The f i r s t says t h a t m L A and rn L 3 are exactly the same number. The second says that the distance A 3 and the distance CD are exactly the same number. Each of the congruences on the right I s a congruence between geometric fi~ures. We do n o t write = between two geometric figures unless we mean that the figures are exactly the same, and t o i n d i c a t e that t h e segments
occasions when we mean t h i s are rare.
One example Is this:
Here it is correct to write L B A C = LEAD,
because L 3 A C and LEAD are n o t merely congruent, they are ex,actly -the same angle. Similarly, and are always exactly the same segment, and so it is c o r r e c t to write AB =
- x.
-
Consider now a correspondence ABC
DEF
between t h e vertices of two triangles AABC
and ADEF.
This automatically gives us a correspondence betww.n t h e s i d e s of
t h e t r i a n g l e s , like this:
-
AB
"EE
and it gives us a correspondence between the angles of the two triangles, l i k e t h i s : LA Z-D LBLE LC LF.
-
We can now s t a t e t h e d e f i n i t i o n of a congruence between two triangles.
-
Given a correspondence ABC DEF between t h e vertices of two triangles. If every pair o f corresponding sides are congruent, and every pair of corresponding angles are congruent, then the correspondence AX-DEF is a congruence between -the two triangles You should read t h i s d e f i n i t i o n a t least twice, very c a r e f u l l y , t o make sure t h a t it says what a d e f i n i t i o n of the idea of a congruence between triangles ought to say. There is a shorthand f o r writing congruences between triangles. When we write L A 2 ZD, t h i s means t h a t the two angles LA and LD are congruent. (That is, m L A = m L D.) Similarly, when we write
Definition.
.
AABC 2
ADEF,
-
this means that the correspondence
DEF is a congruence. Notice that this is a very efficient shorthand: t h e single expression AAEC = ADEF t e l l s us six things a t once; ABC f\s
namely , AB = DE
~~~
AC =
I^^F
BC
DF
= EP
S?3?^
2 ZD Z B 2 LE "^ m Z C = mdF. L C = LF. In each of t h e s e s i x l i n e s , t h e equations on the left and the congruences on the r i g h t mean the same thing, and we can choose either notation at any time, according to convenience, Usually we will write AB = DE, instead of 2 'GE, simply because it is easier to write. For the same reason, we will usually write L A ~ Z Di n s t e a d of m L A = m L D , mZA = m Z D mLB = m L E
L A
It is sometimes convenient to i n d i c a t e a congruence graph1 c a l l y by malting marks on the corresponding sides and angles, like t h i s :
ADEF We can also use t h i s method t o i n d i c a t e t h a t c e r t a i n corresponding parts of two figures are congruent, whether or not we know about o t h e r parts. AABC 2
The marks in t h e figure I n d i c a t e t h a t (1) A 3 = DE, (2) AC = DF and ( 3 ) m L A = m Z D . Question: Would It be correct to write A 3 2 DE, or L A = AD? Why o r why n o t ? It seems pretty clear, in the above figure, that the congruences we have i n d i c a t e d are enough t o guarantee t h a t the correspondence ABC-DEF Is a congruence. That is, if these t h r e e p a i r s of corresponding parts are congruent, the triangles must also be congruent. In f a c t , t h i s is the content of t h e basic congruence postulate, to be stated in t h e next section.
Problem Set
A ABF
a AKiRQ.
Complete the following list by telling what should go In the blanks. A6 M-Q
1;. .
2
AFBR. L i s t the six pains of corresponding, congruent part s of these two triangles. Ar"M 2 AFHW. L i s t the s i x pairs of corresponding, congruent parts of the-ie triangles. (it is n o t necessary t o have a picture but you may make a sketch if you w i s h . ) ARQP 2 AABX. List t h e s i x pairs of corresponding congruent AABR
3.
-
p a r t s of t h e s e triangles.
Do not use a f i g u r e .
2
L i s t the six pairs of corresponding, congruent parts of these t r i a n g l e s , Here is a list of the six pairs of corresponding parts o f two congruent triangles. Give the names of the two triangles t h a t would fit In the blanks below. -m,L A 2 LM. AB = MK. --" BW = T3?. ZBZ LK. AAZW
ABZW.
m ^ m. A.
Z W S ZF. 2 -
A-â
If AABC 2 AXYZ and ADEF 2 AXYZ, what can be said about the relationship of AAEC to ADEF? State a theorem generalizing t h i s situation.
Using ruler and protractor, draw a triangle ABC in which A 3 Is 3 inches long, EC Is 2 Inches long and angle B Is 50'. Compare your triangle with those of other mem-
bers of the c l a s s . Draw AABC In which AC is 3 inches long, BC is 4 inches long and angle C I s . '07 Compare t r i a n g l e s . Draw AABC with A B 3 inches l o n g and EC 2 inches long. Make d B any size t h a t s u i t s your fancy. Compare triangles. If these three exercises suggest t o you an Idea concerning a congruence between two triangles, try to s t a t e or write t h i s idea f o r t r i a n g l e s In general.
9
a.
Given that AABC and ADEF do not i n t e r s e c t , and t h a t X is a p o i n t between B and C , Tell which of t h e syrnb o l s =, = may be i i l l e d in t h e blanks to make t h e s t a t e ments meaningful and possibly true. 1\f
1.
AB - DE . EC - EF . m L A
v
3. 4.
5. 6. 7,
b.
5-3,
^3-
LC.
LAX. m Z ABX rn .L EDF, Which of t h e .blanks could have been filled with either
or 2 c.
- ADEF. -m L D .
AABC
2.
LAEX-
-
=
-
?
had been t h e same segment as DE but if C were a d i f f e r e n t point than F, which blank c o u l d have been f i l l e d by = t h a t should otherwise have been filled by 2 ?
If
B s I c Congruence Postulate.
To get a t the facts on congruences of triangles, we need one new postulate. In t h e name of t h i s postulate, S.A.S. stands for Side Angle S i d e .
s.
Postulate h he S , A .S, P o s t u l a t e . 1 Given a correspondence between t w o triangles ( o r between a triangle *and i t s e l f ) . If two sides and t h e inc luded angle of t h e first triangle are congruent to the corresponding parts of t h e second triangle, then the correspondence is a congruence.
To Illustrate this, we r e p e a t t h e previous f i g u r e .
The postulate means t h a t if
---
AB = DE,
AC 2
and L A
2 ZS,
as i n d i c a t e d in the figure, then
AABC 2 ADEF; t h a t is, t h e correspondence AX-DEF is a congruence. It is very important t o n o t i c e t h a t in the S.A.S. P o s t u l a t e , t h e given angle is the angle i n c l u d e d between the t w o g i v e n s i d e s , like this:
Postulate says t h a t t h e correspondence ABC-DEF is a congruence. If we knew merely t h a t some one angle and some t w o s i d e s of t h e f i r s t t r i a n g l e were congruent t o the corresponding p a r t s o f the second t r i a n g l e , then it would not necessarily follow t h a t the correspondence was a congruence. For example, consider t h i s figure; Under these conditions, t h e S . A . S .
Here AB = DE, L A '= f D , BC = E F , Note t h a t LA and LD a r e n o t included by t h e pairs o f congruent s i d e s . This correspondence is c e r t a i n l y n o t a congruence, because it matches AC w i t h E,
L B with ZE, and
LC with
LF. Since these are not congruences,
the definition of congruence between triangles I s not s a t i s f i e d .
5-4. Writin6 -Your Own P r o o f s , You now have enough basic material to be able t o write real geometric proofs of your own. From now on, writing your own proofs will be a very important part of your work, and t h e chances are that I t will be more fun t h a n reading o t h e r people's proofs. L e t us take a couple of examples, to suggest how we go about f i n d i n g proofs and writing them up. Example 1. If two segments b i s e c t each o t h e r , t h e segments joini n g the ends of t h e given segments are congruent. Given; AR and bisect each o t h e r at F.
-
'
B
S t a r t i n g t o work on a problem l i k e this, we should first draw a f i g u r e and l e t t e r I t , using a capital letter f o r each v e r t e x . Then, s t a t e t h e h y p o t h e s i s and conclusion In terms of t h e l e t t e r i n g of t h e figure Next, we divide t h e page into t w o columns as shown, and write In t h e headings Statements and Reasons, A l l t h i s , o f c o u r s e , Isn't going t o do us a b i t of good unless we can t h i n k of a proof to w r i t e down. Since o u r o b j e c t is t o prove two segments congruent, we must r e c a l l what we know a b o u t congruent segments. Looking back we c a n f i n d t h e definition of congruent segments, of congruent triangles, and the S.A.S. P o s t u l a t e . These are t h e available weapons about congruent segments in o u r arsenal, and a t t h i s p o i n t t h e search i s s h o r t , because o u r arsenal I s small.
.
To apply the postulate, we have to set up a correspondence between two t r i a n g l e s , in such a way that t w o sides and t h e Included angle of the f i r s t triangle are congruent to t h e corresponding parts of the second t r i a n g l e . From the figure, this correspondence l o o k s as if I t ought to be AFB-RFH. Two p a i r s of sides are congruent, because we have f r o m the given data and t h e d e f i n i t i o n of b i s e c t that' AF = RF and J3.F' = HF. How about the included a n g l e s , LAFB and ZRPH? We need t o know that they are congruent, too. And they are, because they are vertical angles. Therefore, by t h e S . A . S P o s t u l a t e , o u r correspondence I s a congruence. The sides and are corresponding sides, and so t h e y are congruent. This I s what we wanted t o prove. Written down in tine double-column form, our proof would l o o k l i k e this: Given: AR and bisect each o t h e r at F. A --"TO prove: AB = m. +R 0
.
-
Reasons
Statements 1. 2.
3.
A F = RF BF=HF. ZAFB 2 LRF'H.
2.
D e f i n i t i o n of b i s e c t . Definition of bisect.
3.
Vertical angles are
1.
congruent.
4.
The S.A.S.
Postulate. 5. Definition of a congruence 5. between triangles. This i s given merely as a sample of how your work might look. There is a limit to how standard1 we can expect t h e form of a proof to be. For example, in t h i s proof we have Indicated congruences between segments by writing AF = HP and BF = HF, and ~ t .
A A F B ~ARPH.
-*AB = RH.
- + -
---
Ve c o u l d j u s t as well have written AF = RF, BF = HF, and so on, because in each case t h e congruence between the segments and t h e equation between the distances mean t h e same thing. so on.
There a r e only t w o really important t h i n g s In writing p r o o f s , F i r s t , what you write should say what you r e a l l y mean. Second, t h e things t h a t you really mean should form a complete l o g i c a l explanation of why t h e theorem I s true, By now, you should have t h e idea, and so we give we second example in an incomplete form. Your problem I s to f i l l in t h e blank: spaces in such a way as to get a proof. H Example 2. -*-
Given:
AH- = FH. HB b i s e c t s LAW. To prove: L A 2 LF. -
Reasons
Statements 1.
-*AH=PH..
1. 2.
Given. Definition of the bisector
of an angle. 3.
Every segment is congruent to Itself.
4.
5. o f t e n made.In proofs is that t h e student assumes as true t h e very thing he is t r y i n g t o prove to be true. Another common mistake is to use a s a reason In his proof a theorem which is actually a consequence of the fact that he is trying to prove. Such arguments are called circular arguments, and are worthless as logical proofs. A p a r t i c u l a r l y bad kind of c i r c u l a r argument is the use of the theorem we are trying t o prove a s a reason f o r one of the steps In its p r o o f " . A mistake
--
Problem S e t 5-k (Note: I n some of the following problems we make use of a square. A square ABCD is a plane figure that is t h e union of f o u r congruent segments AB, 3,E,DA such that L A E C , Z E D , ZCDA, ^ D A B are right angles. Tile square will be discussed in a later chapter of the t e x t . ) 1. In each pair of t r i a n g l e s , if l i k e markings i n d i c a t e congruent parts, which triangles could be proved congruent by S.A.S.?
-
-
2.
In the figure I t is given that intersects "¤ at C, t h a t AC = DC and BC = EC. Show (i.e , prove) that ZB 2 LE. Copy t h e followB ing proof and supply t h e missing reasons.
.
Reasons
Statements
2.
AC=CD. 3C = EC.
3.
ZACB
4.
AACB 2'
1.
2
E
1. LEE.
3.
ADZE.
4.
5.
Given.
angles a r e
congruent .
.
[ ~ o t ethat Statement 3 refers to angles and Statement 4 to triangles, so t h a t your reason here should refer to triangles], Corresponding partsof congruent t r i a n g l e s a r e .
Suppose In t h i s figure L x 2 f y and 3 is the midpoint of AF.
mzm,
show t h a t
-
ZR
ZH. A
Copy and complete t h e following p r o o f .
F
Reasons
Statements 1. 2. 3.
4.
B
I
"-
Given.
From the definition of rnidpoint. S.A.S.
a . If ABCD I s a square and R is the midpoint of AB, prove that RC = RD. ( s e e
-
note preceding Problem 1,) What p a i r s of congruent acute angles appear i n the figure? Prove your answer. In this figure A 3 = FH and m A x = m Lg. Show that m Z A = m L F. b.
A
In this f i g u r e it is given that m Z A E H = m .L FBH, A 3 = FB. Prove AH = F'H. Prove that if segments then A F A B Z AFHR.
AH, 1?B
A
b i s e c t each other at p o i n t
Prove: If the line segments AD and "S5" bisect each other, then
A3 =
DC
and
AC =
DB.
a. Given: Square ABCD, R I s t h e midpoint of F Is a p o i n t between A and D, Q is a p o i n t between C and 3, DF = CQ. To prove: RF = RQ.
m,
A r e there two o t h e r p o i n t s F', Q1 of square ABCD n o t on AD o r such t h a t RF' = RQ'? Where are they? b.
-
Suppose in this figure t h a t AB = AH and that A? b i s e c t s ZHAB. Prove t h a t FH = FB.
A
F,
Overlapping Triangles. Using the Figure & Statements. Frequently In geometric figures, the triangles that we need t o work with are not entirely separate but overlap, like' AAFM and AFAH in the figure below. R
5-5.
The e a s i e s t way to avoid getting mixed u p , and making mistakes, In dealing with such cases, is to write down congruences i n a standard form, l i k e t h i s , AAFM
2
AFAH.
Check t h a t the correspondence AFM-FAH r e a l l y is a congruence, and then later refer back to AAFM AFAH when we want to conclude that two corresponding sides (or corresponding angles) are
congruent. Of course, if you don't see t h e congruences between the overlapping triangles, you will have nothing to check and nothing to apply later. To practice up, write all t h e congruences t h a t you can between triangles contained in the figure above, if it is given that AR = FR and M, H, B are the midpoints o f the respective s i d e s . L e t us now l o o k at a case In which this s o r t of thing comes up in t h e proof of a theorem. Given: HA = HI?. HM = HQ. To prove; PM = AQ.
A
F
very common way to prove that two segments are congruent is to show t h a t the segments are corresponding s i d e s of congruent triangles! If t h i s way can be used successfully here, then t h e f i r s t thing to do is locate the triangles which contain FM and AQ. These are AHMF and AHQA, and these triangles overlap quite a bit. Now the problem becomes one of proving the triangles congruent. The proof i n the double-column form goes l i k e this: A
S t a t ement s
1. 2.
HA = HP. LH 2 ZH.
1
Reasons 1. Given. 2. An angle is congruent to itself.
4.
KM=HQ. A H M F ~ AHQA.
3. 4.
why?
5.
FM
5.
Why?
3.
= AQ.
Why?
s t r i c t l y l o g i c a l proof must not depend on a figure but must follow from the postulates, the d e f i n i t i o n s , and the previously proved theorems. But geometers I n p r a c t i c e use f i g u r e s as a matter o f convenience, and readily accept many observable facts without a tedious restatement in words, unless such a restatement i s essential t o c l a r i f y i n g the problem a t hand. To i l l u s t r a t e , let us l o o k at a restatement of Example 1 used A
previously. Example 1. Le.t A, B, F, H and R be f i v e non-collinear p o i n t s lying in a plane. If (1) F is between A and R, ( 2 ) F I s between 3 and H, (3) AF = FR, and ( 4 ) BF = FH, t h e n (5) AB = RH. T h i s conveys all the information conveyed by t h e figure on t h e l e f t and t h e notation on t h e right below.
AR and
b i s e c t each other at F. To prove: A 2 T3T. Given:
Notice t h a t (1) t e l l s us t h a t and $ are opposite rays, and (2) t e l l s us t h a t % and FH are o p p o s i t e rays. These two things, taken t o g e t h e r , m e a n that LAFB and LRFH are v e r t i c a l angles. (See definition of vertical a n g l e s . ) This I s t h e sort of information t h a t we normally read from a figure. In s t a t i n g problems In t h i s t e x t we will frequently avoid tedious r e p e t i t i o n by referring to a figure. You can use the f i g ure to give the colllnearlty of points, t h e order of points on a l i n e , t h e location of a p o i n t in the Interior o r e x t e r i o r of an angle or in a certain half-plane, and, i n general, t h e r e l a t i v e position o f points, l i n e s , and planes. Things you cannot assume because " t h e y look t h a t way" to you are the congruence of segments or angles, that a c e r t a i n point is a midpoint of a segment, that two l i n e s are perpendicular, nor that two angles are complementary.
-
1.. If In this figure AC = DB, L A C F 2 ZDBE and PC = EB, prove t h a t AF = DE. 2. In t h i s f i g u r e BC = ED AC = AD and LACE 2 LADB. Prove AACE ~ D B .
Proof:
B C
A
(~111in the blanks. )
3.
Statements BC = ED. CD = DC. BD = EC.
4.
AC = AD.
5
LACE
1. 2.
A
2
LADB.
D 1. 2.
Given.
3.
Addition, from statements 1 and 2 .
E
Prove t h a t the diagonals of a square are of equal l e n g t h . (see n o t e preceding Problem 1
of Problem Set 5-4 .) Given: ABFH is a square. To prove: A F = BH.
In t h i s figure LAW 2 LRHQ and F is the midpoint of m. Can you prove AWBP 2 AQHF? Explain.
-
If A B F H is a square and AX, "B? are congruent segments on ->
->
the rays AH, BF respectively, show t h a t are congruent. Restatement : Given: ABFH is a square. + + X, Y are p o i n t s of AH, BF, respectively.
m,
^ Z W
To prove: -AY* =- EX. In the f i g u r e , X is between A and H, and Y is between B and F. Would t h e proof be affected if H were between A and X, and F were between B and Y? Suppose I t 1s given In this f i g u r e that AH = BF, r = m and x = y . Prove that HB = FA.
7
a
If t h e figure B R L AR~ =, RX In
prove t h a t
1T5?,
and ER = RY, A 3 = XY.
me I&osceles T r i a l @ Theorem. -
The A n ~ l eBisector Theorem. A t t h e end of Section 5-1 we mentioned the case of matching up the v e r t i c e s of a t r i a n g l e AABC in which at least two sides of t h e triangle are of t h e same length. This, i n f a c t , i s the case t h a t we deal with In our f i r s t formally stated congruence theorem: Theorem 5-2. If two sides o f a triangle a r e congruent, t h e n the angles opposite these s i d e s are congruent. Restatement: Given a t r i a n g l e AABC. If AB = AC, then LB 2 L C .
5-6.
-
Proof : Consider t h e correspondence ABC ACB, between AABC and Itself. Under this correspondence, we see that
^B-AC,
A5'-m,
-
LA LA. Thus two sides and the Included angle of AABC are congruent to the parts t h a t correspond to them. By the S.A.S. Postulate, this means that AABC 2 AACB, that is, t h e correspondence ABC-ACB I s a congruence. By the d e f i n i t i o n of a congruence between triangles all pairs of corresponding parts a r e congruent. Therefore ZB 2 LC, because these angles are corresponding parts. We now show how t h e above proof looks in t w o - c o l m form. The same figure Is used.
Theorem 5-2, If two s i d e s of a triangle are congruent, then t h e angles opposite these sides a r e congruent. -*Given : AABC with AB = AC. To prove: L B Z LC. Proof: Statements Reasons -.%1. Given 1. AB = AC.
AC 2 AB.
2.
3.
4.
f\f
L A =
AABC
LA.
2
AACB.
ZB 2 L C .
2. 3.
I d e n t i t y congruence. Steps 1 and 2 and the S.A.S.
4.
Postulate. Definition of a congruence between triangles.
Usually, we will state theorems in words, as we have stated Theorem 5-2, and then restate them, using notation which will be the n o t a t i o n of t h e p r o o f , Definitions. A triangle w i t h t w o congruent s i d e s is called isosceles. The remaining s i d e is the base. The two angles t h a t i n c l u d e the base a r e base angles. In these terms, we can state Theorem 5-2 in this form: ' T h e base angles of an i s o s c e l e s triangle are congruent. " Definitions. A triangle whose three s i d e s are congruent is called e q u i l a t e r a l . A t r i a n g l e no two of whose s i d e s are congrue n t is called scalene. Definition. A triangle is equiangular if a l l three of its angles are congruent. Using the term equiangular we s t a t e a theorem which r e a d i l y f o l l o w s from Theorem 5-2. We denote t h i s theorem as Corollary 5-2-1. A corollary Is a theorem which is an easy consequence of a n o t h e r theorem. The proof of Corollary 5-2-1 is l e f t f o r you t o
-
do.
Corollary 5-2-1. Every equilateral triangle I s equiangular, In proving theorems f o r yourself, you will need to make your own figures. It is important to draw figures in such a way that they remind you of w h a t you know, without suggesting more than you know. For example, t h e figure given in the proof of Theorem 5-2
l o o k s l i k e an Isosceles triangle, a n d this I s as it should be, because t h e hypothesis of t h e theorem says t h a t t h e triangle h a s two congruent s i d e s . In the f i g u r e for the S.A.S. P o s t u l a t e , i t looks a s if AABC = ADEF, and this I s as I t should be, because this is t h e situation dealt with In the p o s t u l a t e . But it would not have been good t o draw isosceles triangles to illustrate t h e S.A.S. P o s t u l a t e , because this would suggest t h i n g s t h a t the postulate doesn't say. + Definition. A ray AD b i s e c t s , or is a b i s e c t o r of, an angle ZBAC if D is in the i n t e r i o r of LBAC, and Z3AD 2 LDAC. f\f
Ñ
Note t h a t if AD
B
bisects
LBAC,
then m LBAD = m LDAC
Theorem 5-3. Every angle has e x a c t l y one b i s e c t o r . Proof: Given L A . By t h e Point P l o t t i n g Theorem we can find and C, p o i n t s on t h e sides of L A , such that (1) AB = AC.
L e t D be the mid-point of BC, so that (2) DB = DC. Since AB = AC, I t follows by Theorem 5-2 that (3) LB 2 ' LC, h his follows even though the Isosceles triangle AABC is " l y i n g on i t s s i d e . " ) From (l), (2) and ( 3 ) , and t h e S.A.S. Postulate it
follows that
AABD 2 AACD.
T h e r e f o r e , ZBAD 2 ZCAD, and so m ^BAD = m Z C A D , By t h e def'ln i t i o n of bisector of an angle, this means t h a t & b i s e c t s LBAC. To j u s t i f y our use of t h e word "exactly" we must prove that 3 1s t h e only r a y h a v i n g t h l s p r o p e r t y . Suppose there is a ray A^ a l s o a b i s e c t o r of LA. Then m Z C A D = m ZCAE, s i n c e each of J. these equals à rn LBAC Applying the Angle Construction Postuc 0 l a t e to t h e half-plane with AC as edge shows that we m u s t have that Is, and stand f o r t h e same ray. Hence, t h e r e is exactly one b i s e c t o r . The following definitions are useful I n discussing p r o p e r t i e s of triangles. D e f i n i t i o n . A median of-a triangle is a segment whose endp o i n t s are one vertex of t h e triangle and the mid-point of the opposite s i d e . I k f 3 n i t i o n . An angle b i s e c t o r of a triangle is a segnent
.
3
~3,
3
8
--
whose end-points are one vertex of the t r i a n g l e and a p o i n t of the opposite side which lies in the ray bisecting the angle at the given v e r t e x . Note t h a t every triangle has three medians and t h r e e angle b i s e c t o r s . The figure shows one median and one angle b i s e c t o r of
t h e median from
B, and
is t h e angle b i s e c t o r from Problem Set 5-6
1.
I n t h e f i g u r e A B = A C . We s t a r t t h e proof t h a t Zrn = Ln. complete t h i s proof supplying reasons. f\f
A
B.
Proof :
2.
Zm
Reasons
Statements
is supplementary to
^A=.
l n Is supplementary to Z ACB.
2.
Given: In t h e figure FA = FD and AB = DC. Prove : AAFB 2 ADFC, AFBC 2 AFCB.
3.
If in the figure prove t h a t
c
EB 2 m,
LEEA 2
ZECD.
A 4.
B
C D
If AB = AC and DB = DC in t h e plane f i g u r e , show that LABD 2
LACD.
c 5.
AC = AB and CD = BD t h e plane figure, show
If
LACD 2
ZABD.
In
6.
C
Give a paragraph proof rather than a two-column proof of the following:
X and Y are the midpoints of the congruent sides AC and of the i s o s c e l e s Given:
7. 8.
triangle ABC. A 8 To prove: ZCXY 2 LCYXProve Corollary 5-2-1. v very equilateral triangle Is equiangular. ) A
Given equilateral triangle ABC w i t h Q, R and P, t h e midpoints of the s i d e s as shown. Prove that APQR is equilateral. C
9
5-7,
Prove the following: If median 1^ of to side AB, then AFAB is Isosceles.
-
R B AFAB is perpendicular
-
The Angle S i d e Angle T h e o ~ e m ,
Theorem 5-4.
h he A.S.A.
Given a correspondence between two t r i a n g l e s , (or between a triangle and itself). If two angles and t h e included side of the first t r i a n g l e are congruent t o t h e c o r r e s p o n d i n g parts of the second triangle, then t h e correspondence is a congruence. Restatement: L e t ABC-DSF be a correspondence between two triangles. If L A 2 ZD, AB = DE, then
c
Theorem.)
Proof:
Statements
Reasons
4
1. 1. On the ray DF there is a point Ft such that DF' = AC. 2 . A 3 = DE and m L A = mLD. 2. 3. AAEZ ADEF'. 13* 4. L A B C 2 LDEFt. 4.
The Point P l o t t i n g Theorem. Given. The S.A.S. P o s t u l a t e .
Definition of a congruence between triangles.
I
5.
Given. k and 5, and the definition o f congruent angles Step 6 and p o s t u l a t e 12. # -^* Two lines (EF and DF) intersect in at most one point. Statements 3 and 8.
6. Steps 7. 8. 9.
Ñ
EF
and
.
Ñ
EF1 are the same ray.
F = F ~ .
AABCZ
ADEF,
7.
8. 9.
The p r o o f s of t h e following theorem and corollary are left to the s t u d e n t . The p r o o f s are analogous to those of Theorem 5-2 and Corollary 5-2-1. Theorem 5-5- IF two angles of a t r i a n g l e are congruent, the sides o p p o s i t e these angles a r e congruent. Corollary 5-5-1. An e q u i a n g u l a r triangle I s equilateral.
Problem Set 5-7
In some parts of this exercise there is not enough information to enable you to prove the two triangles are congruent even if you use all other facts that you know, f o r example, that "vertical angles are congruent". If it can be proved that the two triangles are congruent, name the statement { A . S . A . or S.A.S.) supporting 1.
your conclusion; if there is n o t enough information given to prove t h e t r i a n g l e s are congruent, name another p a i r of congruent parts t h a t would enable you to prove them congruent. If there are two possibilities, name both.
134 a.
b. c
.
Given only t h a t AR = MR. A Given only that L A 2 Z M .
f.
Given only that L X F Y 2 d KJ?Y. Given only t h a t L X Y F 2 LKYF.
g.
In accordance with the specifications at the left, list the data which would correctly f i l l the blanks. a. S i d e , angle, side of AABH:
HB.
m. -, b.
Angle, side, angle of AABH:
-'
^s,
-
c.
Angle, s i d e , angle of ABFH:
d.
L HBF. S i d e , angle, side of ABFH:
LF,
-9
7
mj
3.
I=
---
d. e
.
2.
-*-
Given only that AH = AB. Given only that i^c 2 Ld. Given o n l y t h a t L a 2 Z b and LC 2 ^.d.
-*
-9
Follow the d i r e c t i o n s of Problem 2. a. Angle, side, angle of AABP:
-
BF, Side, angle, s i d e of ARAF:
-' b. c
.
-'
ZR, Side, angle, side of ARAB:
-,
d. e.
f
.
m,
L B , -*
Side, angle, side of ARAB:
m.
-9
Angle, side, angle of ARAF: LR,
-9
L. RFA.
Angle, side, angle of AAFB: ZFAB,
3, -.
A
A
F
Follow t h e d i r e c t i o n s o f Problem 2. a. Side, angle, s i d e of AHFB:
-.
-' b.
ZHW, Angle, side, angle o f AABH:
-, c.
d.
HB, Side, angle, s i d e of AHFB:
HE,
-m
w.
__>
-
Angle, side, angle o f AHFB:
-' e.
-
3F, -* Side, angle, s i d e of AABH:
-
AH9
-3
AB.
If bisects and L a 2 Lb in the f i g u r e , prove that cS? bisects
z.
pLV;
G A
Prove Theorem 5-5.(If two angles of a triangle are congruent, the sides opposite these angles are congruent.) Restatement: If I n AABC, L B 2 &, then AB = AC.
Hint: Use congruency of t h e triangle w i t h I t s e l f . B
C
Prove Corollary 5-5-1. v very equiangular triangle is equilateral ) U s e a paragraph proof.
.
If AABC is equilateral, prove AAEC 2 ACAB. If t h e b i s e c t o r of Z G In A FGH Is perpendicular to the o p p o s i t e side a t K, then t r i a n g l e FGH is isosceles. Given:
The figure with f^ Z x = Z y and H3 HP 2 HR.
^ m.
Prove;
136 11.
12.
ffi
I n t h e figure, bisects LRMS and LRWK 2 LSWK. Can it be proved that Z R 2 ZS? If so, do so, Prove that '\1
LA =
LR
AN
2
and Z x
if dy
R
m,
2 in
the figure.
"13. a .
I f , in t h e figure, X the midpoint of AB,
R
is
LA 2 Z B and LAXR 2 LBXF, , show t h a t 2 -Do you need as a part of the hypothesis that the figure lies in a plane?
m.
b.
14.
x
A
B
La = Z b and Z w 2 Zs in the figure. -wProve: GR = KH.
Given:
H
G Can the following be proved on t h e basis of the informat i o n given? Given: LAOB with OA = OB
P, Q, points on rays 66 w i t h AQ = BP. Prove: OP = 00.
and
a,
A:
0
B
0
Prove t h a t RX = RY if it is given t h a t In the figure: BQ = TS, m d B = m LT and m -^Q = m L S .
S
Y
T
5-8-
The Side Side Side Theorem. ----
.
Theorem 5-6. ( ~ h eS. S S, Theorem. ) Given a correspondence between two triangles (or between a triangle and i t s e l f ) . If all three pairs of corresponding s i d e s are congruent, then the correspondence is a congruence. Restatement : Let A B C -DEF be a correspondence between two triangles. If A 3 = DE, AC =
BG 8
Proof: 1.
2.
AABC
then
=
DF, EF,
2 ADEF.
Reasons
Statements 4
1. There I s a r a y AG such that L C A G s- LFDE, and such t h a t G are on opposl'te side of a^. 4 2. There is a p o i n t E1 on AG
such t h a t
AEt =
s
The Angle Construction Postulate. The Point Plotting Theorem.
DE. 3.
The S.A.St
What we have done, so far, is to duplicate
under side of AABC, using the S.A.S.
AB = DE
and
t h e line
'E1?1
intersects H.
^3' In a p o i n t
ADEF on t h e
Postulate. 4.
6 . The segment
Postulate.
by hypothesis; DE = AE1, from State-
5.
ment 2 , BC = EF, by hypothesis; and EF = E t C from S t a t e -
6.
ment 3. By Statement 1, 3 and E 1 are on opp I t e a i d e s of the line
a.
We shall now complete the proof for t h e case in which H I s between A and C, as in the figure. The other possible c a s e s will be discussed later. 7 . Statement 4 and Theorem 5-2. 7. LABH 2 L A E I H . 8. ZCBH 2 LCEtH. 8. Statement 5 and Theorem 5-2. 9 . m L A B H + mLCBH = m L A B C . 9 . The Angle Addition Postulate. 10. m f A E ' H + m^CETH = m LAETC. 10. The Angle Addition Postulate. 11. . L A X 2 Z A E I C . 11. Statements 7, 8, 9 and 10. 12. Statements 3 and 11, 12. L A B C ZDEF. 13. AABG 2 ADEF. 13. Statement 12, the hypothesis, and the S.A.S. Postulate.
This completes the proof for the case in which H is between <-> A and C. We recall that H is t h e point in which the l i n e BE1 <-Ã intersects the l i n e AC, If H = A, t h e n B, A and E t are c o l linear, and the f i g u r e looks l i k e t h i s :
In this case Z B 2 Z E r because the base angles of an isosceles triangle are congruent. Therefore dB 2 LE, "because LE 2 LET. The S.A.S. P o s t u l a t e applies, as before, to show that A A X 2 ADEF. If A is between
H and
C,
then the figure looks l i k e t h i s :
and we show that ZABG 2 LE by subtracting the measures of angles, instead of by adding them. That is,
- mLHI3A m LAErC = m ZHEIC - m ZHEtA,
m L A E = m L W
and so t h a t LAW, LAEtC 2 LDEF, as b e f o r e . The r e s t of the proof I s exactly the same as in the f i r st case
.
The two remaining cases, are similar to the two above.
H
=
C
and
Pmblem S e t 5-8
1.
Given: AABF and AAHP with AHSm and H?zm. Prove:
2.
LHAFZ ZBAF.
In t h e figure, 2 i% and f\i AH = FB. Show that ^.P = L s .
-+-
3.
In the figure, z m and ^? 2 E . Prove t h a t Z H 2 Z R .
C
between
.
A
A
and
H,
Ã
4.
Consider the pairs of triangles pictured below. If on the basis of our information t o date they c a n be proved congruent, t e l l which congruency statement you would use.
m xa b
too"
A
d.
A
Consider
ARWMand Am.
h.
4
B
B
s
A
ARMW AWMX
i. J.
W
ew5K
AW = XM, AB = XR, /A
Consider:
F
F
and and
7
B
2 /x.
AQMH. AHMK.
xQ
R
W
X
5.
6.
A s u p p l i e r wishes to telegraph a manufacturer f o r some parts
in the form of triangular metal s h e e t s . In a d d i t i o n to the thickness, kind of metal, and number of pieces wanted, what I s the least he can say in order to s p e c i f y the s i z e and shape of the t r i a n g l e s ? (consider the possibility of more t h a n one c h o i c e . ) Prove t h e following theorem: If the b i s e c t o r of t h e angle opposite the base in an isosceles triangle i n t e r s e c t s the base, I t is perpendicular to the base, A H B Restatement : Given : & A X
with
H a point on ZACH
8.
such that
2 ZBCH.
To prove: 7.
AC = 3C and
mLm.
Prove the theorem t h a t t h e median from t h e vertex of an i s o s c e l e s triangle is t h e b i s e c t o r of the v e r t e x angle.
A
D
B
Prove t h e theorem: The b i s e c t o r of the v e r t e x angle o f an i s o s c e l e s triangle is t h e perpendicular bisector o f the base. Restatement: Given: A ABF w i t h AF = BF and H a point on AB such t h a t b i s e c t s LAFB. To prove : AH 2 and
-
-
mL'.
A
A H
B
9.
a.
Given:
In t h e figure, AF = BR and
-4.-
--"-
AR = BF.
Prove : LAW 2 ZBFR. h he gap in RB was left t h e r e so t h a t the figure would n o t reveal whether or not intersects A F . )
-
-
b.
Do you need as part of the hypothesis t h a t the figure l i e s in a plane?
10. a .
Given: In the f i g u r e , AH = FB, A 3 = FH, and RQ b i s e c t s in K. Prove: OK = RK. Is the figure necessarily
-
planar?
11.
Given: square ABGD with P, Q, R , S t h e midpoints of PC, m, respectively. Prove: A P Q S Z ARQS.
-
m,
m,
12.
Point out why the following argument Is circular, and thereby Invalid.
Theorem:
Given: To prove: Proof: -*1. 2. 3.
4.
5.
AB
=
The base angles of an i s o s c e l e s triangleare congruent. -+A ABC with AB = AC.
0
2 LC. ---
AC.
-AC* =- AB. - 2 -
Reasons
Statements
BC - CB. A A B C S AACB. dB^ LC.
Given.
1. 2. 3.
Given.
4.
S.S.S.
5.
D e f i n i t i o n of congruent triangles.
Identity.
Point out why t h e following argument is c i r c u l a r . Theorem: Given a correspondence between two t r i a n g l e s (or between a triangle and ~ t s e l f ) . If two s i d e s and the Included angle of the f i r s t triangle are congruent to the corresponding parts of the second t r i a n g l e , then t h e correspondence is a congruence. Given: ABC-DEF, B z m , KT=-, LABC 2 LDEP. Prove: A A E Z ADEF.
Statement
Proof:
Reasons
1ADF,
3
be the ray on <Â¥ t h e same s i d e of A 3 as
Let
!&
such that LBAC t % ¥
Angle-Construction I - Postulate.
2.
Ct
is on ray
->
BC, from
s t e p 1.
Given. S t e p s 2 and 3. Given. Step 1. A.S.A.
Corresponding parts. Given Steps 8 and 9. S t e p 10 and the reason f o r S t e p 2. Steps 7 and 11.
.
AABC
^
ADEF.
112.
'Vi
If La = Z b and Lm 2 Lw in the figure, prove that A
KF'TB.
If in t h e figure
Elz
mJJl#
at F, at A, and m / l a = m z; b y can you prove that FB = AB? If so, do so. *
In A HAP, 'points 3 and W a r e on sides T^ and AH, r e s p e c t i v e l y , and = L A ^ , and AW = AB. Prove: rn = HB.
mlm,
A
A
If In he f i g u r e X L m , "S^i b i s e c t s LAW, bisects LAQF and
8
b i s e c t s LF'QR, -*prove t h a t BQ = HQ.
18.
In AABC and AHRW, AB = HR, AC = HW and median 2 median '^S. On t h e basis of theoreins you have had so far, can yov show that AABC 2 AHRIJ?
If so, do so.
H
,.
F
Â¥i
19. Use
t h e diagram f o r Problem 18 and suppose now it is given -*
that AB = HR, EC = RW, and median AF = median prove t h a t AABC 2 AHRW? If so, do so.
*
20.
m.
Can you
Given: P o i n t s A, R, S, and C lie on Line L. R l i e s b e t w e e n A and S. S lies between R and C. 3 and D do n o t l i e on L. AR = CS, AB = CD. BS = DR. a. P r o v e t h a t : L B S A Z LDRC. b. Need the p o i n t s A, R , S, C, B, D be c o p l a n a r ?
*
-
21, In this figure D is the midpoint of AG, BE, and Prove t h a t AEFG 2 A K A .
m.
C
F
-
Does t h e proof for Problem 21 h o l d even if the segments ^5 are not coplanar? 23, Gtven: I n the figure, -.%RQ = SQ. RC 2 SC". Prove that: ZRCA 2 L S C A .
:22.
m,
mlm.
m;
24.
A t r i p o d w i t h three legs of
v
equal lengths VA, VB, VC stands on a plane.
a. W h a t can you say, if anything, about the distances AB, AC, BC? About the six angles O A B , LVAC, LVBA, e t c ? Answer part ( a ) if you are given also t h a t t h e tripod legs make congruent angles with each o t h e r ; t h a t is, L A V B 2 LBVC 2 L A V C . a . Let and b i s e c t each o t h e r at M. Prove t h a t A 3 = RQ and AQ = RB. b. Now l e t ^Z also be bisected at M. How many pairs of congruent segments, as in ( a ) can you f i n d ? c . You probably thought of as l y i n g in t h e same plane as AR and S^. Is t h i s necessary, or do your conclusions In (b) h o l d even if s t i c k s out of the plane of and B^? T r y t o visualize t h e figure in t h e l a t t e r case, and either draw a p i c t u r e or make a model. L e t A A E be any t~tangleand D a point In the plane of t h i s triangle. The s e t consisting of the union of s i x segmerits AB, E,AD, CD we shall call a skeleton o f a tetrahedron. Each of t h e six segments I s called an edge of t h e tetrahedron, each of t h e four p o i n t s A, B, C, D is a vertex, each t r i a n g l e formed by three v e r t i c e s is a face, each angle of a face is a face angle. Edges and faces of a tetrahedron were considered In Problem 11 of Problem Set 3-lc, a. How many f a c e s are there? How many face angles? b. Two edges of a tetrahedron a r e opposite edges if they do not intersect. They are adjacent if they do Intersect. If each p a i r of opposite edges are congruent, are any of the faces congruent? If each p a i r of adjacent edges are congruent, what k i n d of triangles a r e t h e faces? c . Construct an equilateral skeleton of a tetrahedron w i t h t o o t h p i c k s and quick-drying glue or w i t h soda straws by threading string through them. b.
K-
25.
-
26.
- my
m,
[sec. 5-81
Review Problems f o r Chapter
2
Complete: If the v e r t i c e s of two triangles correspond so that every pair of corresponding angles are and every p a i r of corresponding are congruent, then the correspondence is a between the two triangles. Consider t h e s e t of abbreviations A.S.A., S.S.A., S.A.S., S.S.S., A.A.A. a . Which subsets are abbreviations of postulates in this
chapter? b. Which subsets are abbreviations of theorems proved in this chapter? If ARST I s I s o s c e l e s with RT = ST, what correspondences are congruences be tween
the triangle and itself?
Given A F = B F and D F = E F , what would be t h e final reason in the most direct proof that AAFD 2 A m ? That AAEC A m ? 1
In t h e f i g u r e and A F = BH. Prove; R 3 = RF. Given:
6
A
AR =
RH
In the figure f o r Problem 5, if RB t h a t AR = HR.
= Ry
and
AB =
HF, prove
7.
A person wishes to
---
7S?Im.
m,
--
-
-
8.
--
- ---- -
find the distance across a river. He does t h i s by sighting River a tree, T, on the other side opposite a point P, such that Marking the midpoint, M, of he paces a path perpendicular to PQ at Q until he determines the point X where his <-> path meets line TM. What o t h e r segment In the figure has the same l e n g t h as TP? What Is the p r i n c i p a l theorem used in showing tha t: ATPM 2 AXQM ?
7
-
*
-
-
+
\
\\
Napoleon's f o r c e s , marching into enemy territory, came upon a stream whose width they d i d not know. Although t h e engineers were in t h e rear, nevertheless, the impetuous commander demanded of h i s o f f i c e r s the width of the river. A young officer Immediately s t o o d erect on the bank and pulled the v i s o r of his cap down over his eyes u n t i l h i s line of v i s i o n was on the opposite s h o r e , He then turned and sighted along t h e shore and noted the point where h i s visor r e s t e d . He then paced off t h i s distance along the-shore. Was this d i s t a n c e the width of the river? What t w o triangles were congruent? Why?
In ARST: Point X lies between S and T, and SX = SR. Point Q lies between R and T, and SQ b i s e c t s L S , ^X I s drawn Find an angle congruent t o LR, and establish the con- S
-
Given:
m-m,
Lx '='Ly, QB
= WH
and
R
A
The figure with
BLTR,
x
F,
m.
the midpoint of Prove: ABF& 2 A m .
Gtven:
In the f i g u r e , A% = AR and LBAH 2 ZRAH. Prove: F3 = FR.
12.
In this figure given that: A 3 = KF and + RB = RJ?. Prove: AAFR 2 AHBR.
R
13.
Given: In t h e figure, AB = FB and Iffi = RB. Prove : A AQR 2 A FQM.
14. In t h i s figure given that B and F trisect* %B, A Z L - H and A R = H Q . Prove: BW = FW. *Trisect means to separate Into three congruent parts.
-
R
A
F
15. In t h i s figure, given t h a t HA = KB, AF b i s e c t s L HAB and * BF bisects L HBA.
+
Prove:
AF =
BF.
16.
A polygon
17.
If two medians of a triangle are perpendicular to their respective s i d e s , then t h e triangle is equilateral.
ABODE has five sides of equal length and f i v e angles of equal measure. Prove that LDAB 2 LDBA.
Prove:
18. I n t h i s f i g u r e "SB 2 fSS and
^ '\i m* Prove : 2
m m.
LA
2 Z H and
19.
Prove t h a t the b i s e c t o r s of a pair of corresponding angles of two congruent t r i a n g l e s are congruent.
20.
In t h i s figure it
X
is given that:
m
= QR,
~a
Lb,
Z X 2 ZQ. Prove: KA = KM.
21.
In t h i s figure I t is given that Zl 2 Z2, Z3 L4, and JT = JB. Prove: ~5 2 L 6 .
=
22.
If PA
= PB
and QA
= QB
then LAP& 2 ZBPQ. Will t h e same proof h o l d regardless of whether A is in the same plane as P, Q, and 3 ?
a. Prove: If PA = FB, QA = QB and R I s on ?$ as shown I n the figure, then RA = RB.
Must the five points be coplanar? Will t h e proof hold whether or not A is i n the same plane as B, R, P, and Q? b.
In t h i s f i g u r e , points F and H t r l s e c t AT, and points F and 3 trisect MR. If AF = FB, I s A ABT
A MHR?
Prove your
answer. <-Ã
If & RS &
I s perpendicular to each
A
of three different rays,
RA, RB, RC at
R and RA = RB = RC, prove t h a t (Draw your own f i g u r e . )
SA =
SB = SC.
Q
Let A PAB and A &B lie in d i f f e r e n t planes but have the common side AB. L e t A PAB 2 A QAB. Prove that If X is any p o i n t in A 3 then A PQX is isosceles. P
-
Complete Euclidls proof of the theorem t h a t the base angles of an isosceles triangle are congruent. Given: A 3 = AC Prove: L ACB 2 L AX. Construction: Take a point F with B between A and F, and a point H w i t h C between A and H so t h a t AH = A F . Draw
.
CT and BtT.
B
A
*28.
Given: The p l a n e figure ABCD AB = CD, AD = BC. Prove:
AC
and
bisect
each other.
*29. Given:
The figure ABCD w i t h AB = AC, DB = DC, and LBAX 2 L X A Y 2 ^CAY. Prove: AX = AY.
Review Chapters 1 to 5 REVIEW EXERCISES
80.
Follow each with a 11+" or a "-" to i n d i c a t e w h e t h e r you c o n s i d e r the statement t r u e or f a l s e . True will mean "true under all conditionsi1. Every two rays Intersect. AB designates a l i n e . If m £ Q = 100, then L . Q has no complement. A l i n e and a point not on it determine a plane. If a p o i n t I s In the Interior of two angles of a triangle It is in the i n t e r i o r of the triangle. If a line Intersects a plane n o t containing it, then the i n t e r section is one point. The union of t w o h a l f planes is a Whole plane. A p o i n t which belongs to t h e i n t e r i o r of an angle belongs to the angle, If = m, then either A = C or A = D. The intersection of every two half planes Is the i n t e r i o r of
Write the numbers from
1
to
-
an angle. The i n t e r i o r of every triangle is convex. It Is p o s s i b l e t o f i n d two s e t s , neither of them convex, which
have a union which is convex. A ray has two end-points, Experimentation Is always the b e s t way of reaching a v a l i d
conclusion. Given four different p o i n t s , no three of which are collinear, there are exactly six d i f f e r e n t lines determined by pairs of these p o i n t s . If m L R S T = m L X Y Z , then^ R S T ~ ZXYZ. In t h e f i g u r e the best way fco name + +a the angle formed by DA and DC is £D.
In this text b e t w e e n for points on a l i n e I s a n undefined term.
-
The v e r t i c e s of a triangle are non-collinear . The intersection of two s e t s I s the s e t of all elements that belong to one o r b o t h of them. Every statement about geometric f i g u r e s which is not a d e f i n i tion can be proved. If A X Y Z 2 A CAB, then L A 2 L X. It is possible f o r two lines to intersect in such a way that three of the angles formed have measures 20, 70, and 20. . Each side of an angle Is a ray. All nouns w h i c h t h e t e x t uses that relate to geometry are d e f i n e d in the t e x t . The i n t e r i o r of an angle is a convex s e t , If rn L A B C = 37 and m L DEF = 63, t h e n L- A E and i- DEF a r e complementary. If A is n o t between B and C, "then C Is between A and B.
m is never a negative number. If p o i n t Q Is in the exterior o f t - A N , then Q and C are @ on the same s i d e of AB. The distance between two points is the absolute value of t h e sum of t h e i r coordinates. The longest side of any triangle I s called I t s hypotenuse. # ¥
The absolute value of every real number except zero I s positive.
If CD + CE = DE, then D is between C And E. If in A A E , m^. A = m ^.B = m f C, then AB = BC = AC. + + If, in a plane 2, P T l i n~e L, PQL line L, and P 1s on 0 <&Ã L, then PT = PQ. From t h e statements (1) If q is false, then p is false, and (2) p is t r u e , we can conclude t h a t q is true. The Ruler P o s t u l a t e s t a t e s t h a t any unit can be reduced t o Inches.
If R I s a p o i n t in the i n t e r i o r of L X Y Z , then m L XYR + m L Z Y R = rn L XYZ. There are c e r t a i n points on a number scale which are not in correspondence with any number. Every line is a colllnear s e t of points. - n = n. The distance between two points is a positive number. From the facts that m L AOB = 20 and m L BOC 30 it can be concluded that rn L AOC = 50. A p o i n t on t h e edge of a half-plane belongs to t h a t half-plane. A line L in a plane E separates t h e plane i n t o two convex
-
sets.
The median of a triangte b i s e c t s the s i d e to which it is drawn. If two p o i n t s l i e In the same half-plane, t h e n the line d e t e r mined by them does n o t i n t e r s e c t t h e edge of that half -plane. If two supplementary angles are congruent, each is a right angle. The Interior of an angle Includes t h e angle itself. Vertical angles have equal measures. The sides of an angle are rays whose intersection is the vertex of the angle. If L C I s supplementary t o L A and m L A = 67, then m L C = 113, If two lines intersect, there are exactly two p o i n t s of each which are contained by t h e other.
If two angles have equal measures the angles must be congruent. From the statement (1) If p is t r u e , then q Is t r u e , and (2) p is not true, we can conclude that q is false. It h a s been proved In t h e first five chapters of t h i s text t h a t t h e sum of the measures of t h e angles of a triangle is 180. The s i d e s of a triangle are lines. The midpoint of a segment separates it into t w o rays. If two l i n e s Intersect so t h a t the v e r t i c a l angles formed are supplementary, then the measure o f each angle is 90. If m L B = 93, then L I3 is acute. For all numbers x, 1x1 = x. + The intersection of AB and BA is E , In A A B C all p o i n t s of BC are in the interior of L A. If A A B C 2 A BCA, then A A X is equilateral. If 1x1 = j y l , then x 2 = y 2. A ABC and A RFH which are In d i f f e r e n t planes are congruent if AB = RF, PC = FH and AC = RH. A ABC 2 A MQT if AB = QM, BC = TQ and A Q '= L B . Median in AACE b i s e c t s L A . 2 2 IS x = y , then 1x1 = y l .
-
a
If three points are on three d i f f e r e n t l i n e s , the points are There is no A ABC in which L A = L B. Two p o i n t s not on a plane are in opposite half-spaces determined by the plane i f and only if t h e segment joining t h e m intersects the plane.
Chapter 6
A CLOSER LOOK AT PROOF 6-1. How A Deductive System Works. In Chapter 1 we t r i e d to e x p l a i n in general terms how our study of geometry w a s going to work. After the experience that you have had since then, you ought to be In a much better position to understand the explanation. The idea of a s e t , the methods of algebra, and t h e process of l o g i c a l reasoning, are things that we have been working with. The geometry Itself is what we have been working on. We started w i t h point, line and plane as undefined terns; and so f a r , w e have used f i f t e e n postulates. Sometimes, new terms have been defined by an appeal to postulates. (FOP example, t h e distance PQ was defined t o be the positive number given by the Distance Postulate. ) Sometimes definitions have been based only on the example a set of points is c o l l l n e a r if undefined terms. a l l points of the s e t l i e on the same line. 1 But at every polnt we have b u i l t our definitions with terms that were, in some way, previously known. By now we have p i l e d d e f i n i t i o n s on top of each other so often that t h e list is very long. And in f a c t , the length of the list I s one of the main reasons why we had t o be careful, a t the outset, to keep the record straight. In the same way, a l l the statements t h a t we make about geometry are based ultimately on the postulates. Sometimes we have proved theorems d i r e c t l y from postulates, and sometimes we have based our proofs on theorems that were already proved. But i n every case, the chain o f reasoning can be traced backward t o the p o s t u l a t e s . You might find i t a good idea, a t t h i s point, t o reread the second half of Chapter 1. It will seem much c l e a r e r t o you now than i t d i d the f i r s t time. It is much easier to look back, and understand what you have done, than to understand an explanation of what you are about to do. -d
o or
6-2. Indirect Proof. We remarked in Chapter 1 t h a t the best way to learn about l o g i c a l reasoning is by doing some of it. There is one kind of proof, however, that may require some additional discussion. For !heorern 3-1, we used what I s called an Indirect proof. !The theorem and i t s proof were as follows; II Theorem 5-1. Two different lines intersect in at most one point. Proof: It is Impossible for two different lines to intersect i n two d i f f e r e n t p o i n t s P and Q. This is impossible because by Postulate 1 there is only one l i n e that contains both P and Q." Probably this was the first time that you had seen this kind o f reasoning used in mathematics, but you must have encountered t h e same s o r t of thing, many times, in ordinary conversation. Both of the following remarks are examples of indirect proofs; (1) "1t must be raining outside. If it were not raining, then t h o s e people corning in the door would be dry, but they a r e soaking wet. " (2) "Today must not be the r i g h t day f o r the f o o t b a l l game. If the game were today, then the stadium would be full of people, but you and I are the o n l y ones here. It In each of these cases, the speaker wants t o show t h a t his f i r s t statement is true. He starts his proof by supposing t h a t the thing he wants t o prove is wrong; and then he observes that this leads to a conclusion which contradicts a known fact. In t h e first case, the supposition is that it is not raining; this leads to t h e conclusion t h a t the people coming in would be dry, which contradicts t h e known f a c t that these people are w e t ; and
therefore it is raining, after all. Similarly, in the second case the assumption that the game Is today l e a d s t o a contradiction of t h e known fact of the empty stadium. In the proof of Theorem 3-1, the supposition is that some two d i f f e r e n t lines i n t e r s e c t in two points. By Postulate 1, t h i s leads to the conclusion that the l i n e s aren't different a f t e r a l l . Therefore t h e supposition is wrong, and this means
that the theorem I s right.
Problem S e t ---. 6-2a
For the sake of argument accept each of the following assumpt i o n s and then give a logical completion for each conclusion. a. Assumption: Only men are c o l o r b l i n d . Conclusion: My mother --------. b. Assumption: A l l men are left-handed. Conclusion: My brother --------. c. Assumption: The only t h i n g that makes Jane ill i s hot chocolate. Jane I s ill. Conclusion: Jane --------. 2, Which of the following arguments are indirect? The temperature outside must be above 32' F. If t h e temperature were not above 32', then the snow would n o t be melting. But it I s melting. Therefore, the temperature must be above 32O. That movie must be very entertaining. If I t were not very e n t e r t a i n i n g , then only a few people would go to see It. But large crowds are going to see it. Therefore, It must be very e n t e r t a i n i n g . The air-conditioning in t h i s building must not be working c o r r e c t l y . If it were working correctly, then the temperature would not be so high. But the temperature is uncomfortably high. Therefore, the air-conditioning i s not working correctly. 3 . Mrs. Adams purchased a s e t of knives, forks, and spoons advertised as a s t a i n l e s s steel product. After u s i n g the s e t for several months, she found that the s e t was beginning to rust. She thereupon decided that the s e t was not stainless s t e e l and returned it f o r refund. In this example of Indirect proof identify (1) the statement to be proved, (2) the supposition made, (3) the conclusion resulting from the supposition, and ( 4 ) the Icnown fact contradictory to ( 5 ) . 1.
What conclusions can in which x , y and z If x is true, then If y I s true, then
you draw from t h e following hypothesis
stand for d i f f e r e n t statements? y I s true. z is t r u e .
x is true. Suppose you have the f o l l o w i n g data: If w is true, then v is true. If u is true, then w is true. If x is true, then u is t r u e v is not true. What conclusions can you draw? Did you use indirect reasoning at any point? What conclusion Follows from t h e following data? (1) Nobody is allowed to j o i n the swimming club unless he can play t h e piccolo. (2) No t u r t l e can play t h e piccolo. ( 3 ) Nobody I s allowed t o wear striped trunks In the club pool unless he is a member of the swimming club. ( 4 ) I always wear s t r i p e d trunks In the c l u b p o o l , ( H i n t : This problem becomes e a s i e r if you convert it to i f - t h e n form, a s in several preceding problems. For example, let A be "someone is a member of the swimming c l u b " , let 3 be "someone can p l a y t h e piccolo", e t c . ) If A is green, then B is red. If A I s blue, then B is black. If I3 is red, then Y is white. a. A is green, so 3 is and Y is b . B Is b l a c k . Is it possible to draw a conclusion conc e r n i n g A? If so, what conclusion? Prove that the bisector of any angle of a scalene t r i a n g l e cannot be perpendicular to t h e opposite side. L e t us now prove the o t h e r theorems of Chapter 3 . For convenience, we first restate the postulates on which these proofs are based:
Postulate 1. Given any t w o different points, there is exactly one line which contains b o t h of them. [ sec. 6-21
1
1 1
Postulate 5. a. Every plane contains a t least t h r e e noncollinear points. b. Space contains a t l e a s t f o u r noncoplanar p o i n t s .
Postulate 6. If two points l i e in a plane, then the line containing these points lies in the same plane. P o s t u l a t e 7. Any three points lie in a t l e a s t one plane, and any three non-colllnear p o i n t s l i e in exactly one plane, More b r i e f l y , any three points are coplanar, and any three n o n - c o l l l n e a r points determine a p l a n e .
s.I f
a line I n t e r s e c t s a plane not containing it, then the intersection is a single point. Proof: By hypothesis, we have a line L and a plane E, and (1) L intersects E In a t least one point P, and ( 2 ) E does not contain L. Theorem
We are going to g i v e an i n d i r e c t proof of the theorem and t h e r e f o r e we start by supposing t h a t the conclusion is false. Thus our supposition is that ( 3 ) L i n t e r s e c t s E In some o t h e r point Q, To give an i n d i r e c t proof, we need to show that our suppos i t i o n contradicts a known fact. And it does: If P and Q lie i n E, it f o l l o w s by Postulate 6 that the l i n e c o n t a i n i n g them lies in E . Therefore ('1) L lies in E, This contradicts (2). Therefore t h e supposition ( 3 ) i s Impossible. Therefore Theorem 3-2 is true.
Notice that t h e figures t h a t we use to Illustrate Indirect proofs look peculiar. In the figure f o r Theorem 3-2, we have indicated a point Q, merely to remind ourselves of the notation of the proof. The proof itself shows t h a t no such point Q can possibly e x i s t . In f a c t , the figures for indirect proofs always l o o k ridiculous, for a good reason: they are pictures of imposs i b l e situations. If we had drawn a figure f o r Theorem 3-1, it would have looked even worse, perhaps like this:
T h i s I s a picture of an impossible situation in which two different lines intersect in t w o different points. Theorem 3-5. Given a line and a point not on the l i n e , there is exactly one plane containing both of them.
L and a point P not on L. By the Ruler Postulate we know t h a t every l i n e contains I n f i n i t e l y many p o i n t s , and so L contains two p o i n t s Q and R. By Postulate 7 there exists a plane E which contains P, Q, and R. Since by Postulate 6, E contains L, we have shown that there exists a plane E containing both L and P. A t this p o i n t we actually have proved o n l y half of the theorem, since Theorem 3-3 says there is exactly one such plane. It remains to prove that no o t h e r plane containing L and P Proof:
By hypothesis we have a l i n e
-
We do t h i s by Indirect proof Suppose t h a t there is another plane E' containing L and P. Since by Postulate 1 L is the only line containing Q and R, we know that Q and R, a s w e l l as P, lie in E* This c o n t r a d i c t s Postulate 7 which says t h a t exactly one plane contains three non-collinear points Since E was established as a plane containing P, Q and R, E1 can not exist, and E is the o n l y plane containing L and P, The t w o p a r t s of the proof of Theorem 5-3 bring up the distinction between existence and uniqueness. The first h a l f o f t h e proof shows t h e existence of a plane E containing L and P. This leaves open the possibility t h a t there may be more t h a n one such plane. The second half of the proof shows the uniqueness of the plane. When we prove existence, we show that t h e r e is a t l e a s t one o b j e c t of a certaln k l n d . When we prove uniqueness we show t h a t there is a t most one. If we prove both existence and uniqueness, t h i s means that there is exactly E. For example, f o r t h e f l e a s on a stray dog, we can usually p r o v e existence, but not uniqueness. (it I s a very lucky dog t h a t has only one f l e a . ) For t h e eldest daughters of a given woman, we can obviously prove uniqueness, but not necessarily existence; some women have no daughters a t a l l . For the points common t o two different segments, we don't necessarily have e i t h e r existence or uniqueness; the Intersection may contain many points, or exactly one point, or no points at all. The phrase o n e and only o n e is of ten used instead oi tr exactly one' since I t emphasizes the double nature of t h e statement. The following theorem breaks up into two parts in exactly exists.
---
t h e same way:
E.
Given two intersecting lines, there 2s exactly Theorem one plane containing them. For variety we give the proof in double-column form. Note the two p a r t s and the way we handle the Indirect proof In the second p a r t .
Proof: We have given the l i n e s in the point P. Statements
L c o n t a i n s a point different from P. Q is not on %. There I s exactly one plane E, containing and Q. E contains L-,.
Q,
5
P contains Q. E and F each contain L2 and Q. E I s the only plane containing Ll and L2.
and
Ln, Intersecting
Reasons
1.
By t h e Ruler
2.
Theorem 3-1. Theorem 3-3.
3. 4.
Suppose t h a t another plane
L,
Postulate, every line contains inf i n i t e l y many p o i n t s .
By Postulate 6 , since contains P and Q.
E
F also contains L, and L2. 6 . Q is on L,. 7. Steps 3 and 4, and 5 and 6. 8.
S t e p 7 contradicts Theorem
3-3.
--
Problem Set 6-2b
Is a t r i a n g l e n e c e s s a r i l y a plane figure? Explain,
Theorem 3-4 says, In effect, "Two intersecting l i n e s determine a plane". How many d i f f e r e n t planes are determined by pairs of intersecting l i n e s in this figure? Assume that the three lines are not a l l in the same
p l a n e . L i s t each plane by naming the two intersecting l i n e s t h a t determine i t . How many d i f f e r e n t planes a r e determined by pairs of the < - Ã ˆ < Â ¥ Ã ˆ * f o u r d i f f e r e n t l i n e s AQ, BQ, CQ, and DO., no three of whlch l i e in the same plane? L i s t the planes by naming f o r each the two intersecting lines that determine it. If, in a plane 2, 1 line L and *"pS -1 line what conclusion can you draw regarding PQ and PT?
5.
3
4.
ÇÃ
and B l i e In plane P. <-> Q lies above plane P. Does l i n e AB l i e entirely in P? Quote a postulate or theorem to support your conclusion. There is a second plane implicit In the situation. Name it by the three points which determine I t . What is the intersection of these two planes? A t what point w i l l <-Ã QB intersect plane F? If A, B, C, D a r e f o u r non-collinear points, list a l l the planes determined by subsets of A, B, C, D. As indicated in this figure,
6.
.
A
Theoxems about Perpendl c u l a r s . Some of the basic theorems about perpendicular lines are good examples of existence, uniqueness, and i n d i r e c t proofs. Theorem 6-1. In a given plane, through a given point of a given line of t h e plane, t h e r e passes one and only one l i n e perpendicular to t h e given l i n e . Given: E is a plane, L a l i n e in E, and P a point of L. To prove: (1) There is a line M In E, such that M contains P and M L; (2) There I s at most one line in E, containing P and perpendicular to L. [sec. 6-31
6-3
1
Proof o f (1):
J>
ti
Y'
-
-
x
P
-
L
IM be one o f t h e two h a l f -plan e s in E t h a t ha as an edge, and let X be a p o i n t of L, d i f f e r e n t from t h e Angle Construction P o s t u l a t e , there is a point Y of H, such that L X P Y 18 a right angle. Let M be t h e l i n e % Then M -L L. Thus we have proved that there 1& at least one l i n e satisfying the conditions of the theorem. Proof of ( 2 ) : We now need to prove t h a t there is a t m o s t one S L C ~line. Suppose that there are two of them, MI and M2. L e t X be a p o i n t of L, d i f f e r e n t from P.
Let
H
--
Then t h e l i n e s M, and Mp contain rays PY., and FY2 l y i n g in t h e same half-plane H having L as its edge. By definition of perpendicular lines, one of the angles determined by L and I s a right angle, and by Theorem 4-8 all four of these M, angles are r i g h t angles. Thus r n f X P Y l = 90. Similarly, mLXPYp = 90. But this contradicts the Angle Construction * Postulate, which says that there I s only one ray PY, w i t h Y in H , such t h a t m L X P Y = 90. This contradiction means that our assumption o f two perpendiculars M, and M2 must be false, which proves t h e second half of the theorem.
The condition 'ln a given p l a n e ' I s an important p a r t of t h e statement of t h i s theorem. If this condition were omitted the first (existence) part of t h e theorem would s t i l l be true but t h e second (uniqueness) part would not. This is e a s i l y seen by thinking of the r e l a t i o n between the spokes of a wheel and t h e axle. Thus l e a v i n g o u t t h i s condition would give us an example of a geometric existence theorem with no corresponding uniqueness theorem. The opposite situation, a uniqueness theorem with no corresponding existence theorem, has already been considered in this chapter. Can you i d e n t i f y i fc? D e f i n i t i o n . The perpendicular bisector of a segment, In a plane, is the l i n e In t h e plane which is perpendicular to the segment and contains t h e m i d - p o i n t . Every segment has exactly one m i d - p o i n t , and through t h e mid-point t h e r e i s exactly one perpendicular line in a given plane. Thus, f o r perpendicular b i s e c t o r s in a g i v e n p l a n e , we have both existence and uniqueness. The following theorem gives a useful characterization of the points of a perpendicular bisector: Theorem 6-2. The perpendicular bisector of a sewenfc, in a plane, i s t h e s e t of all p o i n t s of the plane that are equidistant from the end-points of the segment. Restatement: L e t L be the perpendicular b i s e c t o r o f the segment in a plane E and let C be the mid-point o f
m.
Then
P I s on L, then PA = PB, and (2) If P is in E, and PA = PB, then P is on L. Notice that the restatement makes It plain that the proof of t h e theorem w i l l c o n s i s t of two p a r t s . In the first p a r t we prove t h a t every p o i n t of t h e perpendicular bisector s a t i s f i e s the characterization, that is, is equidistant from t h e e n d - p o i n t s of the segment. But t h e theorem says that the perpendicular bisector I s the set of all such points. To prove this, then, we must also show t h a t every such p o l n t , c h a r a c t e r i z e d by belng equidistant from the end-points of the segment, is on the (1) If
-
perpendicular b i s e c t o r .
This last I s the second p a r t of t h e
restatement. Proof of (1):
Given a point P of L. If P l i e s on AB, t h e n P = C, and this means that PA = PB by t h e d e f i n i t i o n of mid-point of a segment. If P is not on the line AS, then <->
PC
PC by i d e n t i t y , and, by hypothesis, CA = CB and LPCA LPCB. Hence by the S.A.S. Postulate, =
A PCA
S
APCB.
to be proved. Proof of ( 2 ) : Given t h a t P lies in the plane E and PA == PB. If P is on ^S, then P is the mid-point C of m, and so<->P is on L. If P is n o t on E, l e t L * be t h e l i n e PC: Therefore
PA = PB, which was
[sec. 6-31
t h e n PC = P C , Theorem,
C A = C B , and AFCA
=PB.
PA
=
(Why?)
BytheS.S.S.
A PCB.
Therefore L P C A Qt L PCB. Therefore, by definition, L! ( and so L' is t h e perpendicular bisector of Therefore, by Theorem 6-1, L 1 = L, and P is on L, which was to be
m.
proved.
Next we prove the analog of Theorem 6-1 f o r t h e case in which the given point is not on the given line. Since the proof is considerably more complicated than that of Theorem 6-1, we w i l l state and prove the existence and the uniqueness parts as separate theorems. Because it I s the simpler, we s t a r t with uniqueness. Theorem 6-3. Through a given external point there is at most one line perpendicular to a given l i n e . Proof: Like most uniqueness proofs, t h i s is an indirect one. Suppose Li and Lg are distinct lines through point
P, each perpendicular to L.
.
.a
L e t L., intersect L in A and Lg i n t e r s e c t L in 3. Since t h e l i n e s are distinct and both go through P we must have A # B (Theorem 3-1). + On the ray opposite to AP take AQ = AP (Point Plotting he or em). Then AQ = AP, AB = AB, m L P A B = m L Q A B = 90, and so AQAB 2 A PAB by the S.A.S. Postulate. It follows t h a t
<-È
and so BQ J. L. This contradicts Theorem 6-1, which says t h a t there is only one perpendicular t o L a t 3 l y i n g in the plane containing L and Li. Hence our supposition that there could be two perpendiculars t o L through P Is f a l s e . Cp.rollary 6-51. A t most one angle of a triangle can bp a right angle. For If In" A ABC, L A and L B were b o t h right angles we would have two perpendiculars f r o m G to fS. Definitions. A right t r i a n g l e is a t r i a n g l e one o f whose angles is a right angle. The side opposite the right angle is the hypotenuse; t h e s i d e s adjacent to the r i g h t angle a r e the legs.
Theorem 6-4.
Through a given external point there is a t l e a s t one line perpendicular to a given l i n e , Restatement: L e t L be a line, and l e t P be a point not on L. Then there i s a line perpendicular t o L and containing
F i r s t we will e x p l a i n how the perpendicular can a c t u a l l y be constructed, on paper, using a r u l e r and a p r o t r a c t o r . From the
method of construction, it will be c l e a r how t h e theorem can be proved from t h e postulates S t e p 1. k t Q and R be any two points of t h e l i n e L. Measure the angle L PQP.
.
St-ep 2-. Using the prot~actor, construct an angle L RQS, with the same measure a s LPQR, taking S on the opposite side of t h e line L from P. + S t e p 3 . Measure the distance QP, Take a point T on QS, such t h a t QT = QP, <-> Step 4 . Now draw the line TF. This is the perpendicular t h a t we were looking f o r . For the reasons, see the proof below. F i r s t , however, you should try this construction with your ruler and p r o t r a c t o r , and try t o see f o r yourself why It works. Let us now write down the proof In t h e double-column form. Each of t h e f i r s t few statements on t h e l e f t corresponds t o one of t h e t h i n g s that we were doing with our drawing instruments.
Reasons
Statements 1. 2.
3. 4.
L contains two p o i n t s Q and R. There is an angle LRQS, congruent to L RQP, w i t h S and P on d i f f e r e n t sides of L. There is a oint T of the ray such that
The Ruler Postulate.
T and
P
d
s i d e s of
P are on opposite L.
5.
TP
6.
point U, APQU s ATQU.
7.
LQUP 2 L QUT.
8. 9.
intersects
L, in a
The Angle Construction Postulate. The Point Plotting
Theorem. and
S
are on opposite
sides of L, and 3 and T a r e on the same side of Definition of opposite sides. Statement 2, statement 3 , and the S.A.S. Postulate. Definition of a congruence between t r i a n g l e s . D e f i n i t i o n of r i g h t angle. Definition of perpendicularity.
LQUP is a r i g h t a n g l e . PT 1 L.
<-È
This proof somewhat resembles t h e proof of the S.S.S. Theorem h he or em 5-61, Like t h i s e a r l i e r theorem it has sever a l cases, only one of which (that In which U and R lie on t h e same side of Q) is completely covered by t h e above proof. [aec. 6-31
The modifications necessary for* the other t w o cases (U = Q and Q I s between R and TJ) are l e f t as exercises for the student,
t?
Problem S e t J.
6-2
If BC = DC and Is?, prove without the use of congruent triangles that m = ED.
C
E
,
shown in the figure, with lengths o f segments as Indicated, find x, y and
3,
Given: PA = PB, M I s the midpoint of and Q I s ¥
m,
B 4.
Given: The l i n e m is the perpendicular bisector of the segment P I s on the same side of m as Q. R is the i n t e r s e c t i o n of m and "FT. T Prove : PT = PR + RQ.
m.
m
[sec. 6-31
COPY the figure below.
Following the steps outlined in the t e x t construct perpendiculars from A, B and X to l i n e L. 'X
A
B
Using ruler and p r o t r a c t o r construct perpendiculars from A and F t o COPY t h e figure.
Does Theorem 6-4 state the existence of a unique perpendicular to a line from a p o i n t off the line? If we confine o u r t h i n k i n g to a p l a n e , does Theorem 6-1 s t a t e the existence of a unique perpendicular to a l i n e through a p o i n t on the line? Given i s o s c e l e s t r i a n g l e ABC with AC = 3C and b i s e c t o r s and of LA and L B. and BE intersect a t point I?. Prove that C? is perpendicular to (It is n o t necessary to use any congruent triangles in your p r o o f . ) One diagonal of a quadrilateral b i s e c t s two angles of t h e q u a d r i l a t e r a l . Prove that It bisects the other diagonal. In this figure given:
BE
2
RC
=
AD
m.
sc,
Q I s midpoint of L RCA 3
Prove:
SCA.
^&1 RS".
6-4.
Introducing Auxi liary -Sets Into Proofs .
You probably noticed t h a t in proving some theorems, most recently, Theorems 6-2 and 6-4, we Introduced certain points, rays and segments I n t o the figure in addition to those specifled i n the theorem. Possibly two questions concerned you: 1. How can we justify introducing such additional s e t s i n t o p r o o f s on the basis of our postulates? 2 . How do we know which of these sets, If any, should be introduced into the proof of a theorem? The f i r s t question is easy to answer. In working with theorems we usually are concerned w i t h various relationships among certain points, lines, planes and subsets of these, and as a p r a c t i c a l m a t t e r in proving theorems, we choose c e r t a i n planes or l i n e s and certain points on them. Frequently we do n o t concern ourselves with j u s t i f y i n g this procedure. For example, if we are given a l i n e we may immediately name I t When asked to give a reason, however, w e can r e f e r t o t h e Ruler Postulate, which says that a line c o n t a i n s infinitely many points, and thereby t h e two points P and Q exist. S i m i l a r l y , <-Ã given two points A and B we may t a l k about AB with complete confidence since It s t a n d s for a line whose existence and uniqueness are guaranteed by Postulate 1. (See S e c t i o n 6 - 2 . ) The careful concern over Justifying existence and uniqueness becomes especially Important when we i n t r o d u c e into the proof certain points, lines, segments, and so on, not accounted for by t h e theorem being proved. Certainly we can n o t have these sets in our proofs if they do n o t exist under t h e conditions of our geometry, except, of course, In an Indirect proof, where She object is to show they can't exist.
=
In the t a b l e below we list t h e postulates and theorems occurring so f a r which may be used, a p p r o p r i a t e l y , to Introduce auxiliary sets into p r o o f s . Geometric Set
Point. a . Midpoint. Line. Perpendicular
Existence
Postulates 3 and Theorem 2-5. Postulates 1 and
Theorems 2-14, 3-1, 3-2.
Theorem 2-5. Postulates 1 and 8. Theorem 6-1.
at point on line, In a plane. Perpendicular
Uniqueness
he or ems
2-5 and
bisector, in 6-1 a plane. Perpendicular Theorem 6-4. from point
Theorems 2-6 and 6-1. Theorem 6-3
.
not on l i n e , Plane.
Theorems 3-3 and
Postulate 12.
Postulate 12.
Theorem 5-3.
Theorem 5-3.
Postulate 1 and Definition of segment
P o s t u l a t e 1 and Definition of
3-'4. Ray as used in angle measure.
a.
Bisector of an angle.
Segment.
Postulate 7.
Postulate 7 . Theorems 3-3 and
3-4.
segment.
From this table you may see t h a t you a l r e a d y know a l o t about the nature of our three basic undefined terms. The answer to the second question presents a problem quite different from t h e answer to the f i r s t . Getting to know when to introduce a u x i l i a r y s e t s into a proof is l a r g e l y part of the process of l e a r n i n g t o reason logically- It requires considerable p r a c t i c e . Let 's try an example to see how t h i s works.
Example 1. Given: The plane figure with To prove: L D '= LE.
AD
=
AE
and
CD = CE.
A
Since a l l of our p o s t u l a t e s and theorems concerning congruence have d e a l t with triangles, I t seems reasonable that our figure should show some triangles. We can accomplish this e a s i l y
by i n t r o d u c i n g e i t h e r 17 or HE. Suppo8e.we i n t r o d u c e DE so that our figure looks l i k e t h i s :
-
This allows us t o complete the proof, since mLADE = mLAED and m L C D E = m L C E D g i v e s us m L A D C = m L A E C by the Angle Addition Postulate.
Had we Introduced 'KQ Instead of BE, our proof, in twocolumn form this time, would have looked l i k e this:
Proof:
Statements
1.
Introduce TO?.
2.
5.
AC = AC. AD = AE and CD = CE.
4.
A ADC
5.
L-Ds
&AEC.
LE.
Reasons 1. Postulate 1 and D e f i n i t i o n of segment. 2, Identity. 3 . Given. 4. S.S.S. Theorem. 5. Definition of congruent triangles.
Each of the solutions to Example 1 is correct. The choice of which one you use is up to you. But i t i s worth n o t i n g that i n many problems where a choice exists, t h e choice y o u make w i l l determine the degree of d i f f i c u l t y of the p r o o f . It is h e l p f u l t o think through each solution before writing one down formally. An important aspect of learning what t o introduce in a proof can be illustrated if we remove from the hypothesis of Example 1 the condition that the figure is a plane f i g u r e . If D i s not c o p l a n a r with A, E, and C, at least one of the solutions does not hold. Does e i t h e r solution hold? I f one does, which one? One final word of warning before you begin t o Introduce auxiliary s e t s Into y o u r proofs. In answering Question 1 we were careful to say that each such step must be justifiable, that I s , that every p o i n t , l i n e , plane, and so on must exist
under o u r postulates. Students often make the mistake of not recognizing this. For example, you might t h i n k you could prove the statement " ~ l langles are congruentt by the following armment . Example 2, Given any A ABC , prove t h a t L B 3 LC. Proof: In AABC Introduce bisecting L A and perpendicular t o EC'. 0 D C Then L BAD S L C A D by definition o f the bisector o f an angle, AD = AD by Identity, and LBDA a LCDA by t h e d e f i n i t i o n of perpendicular and the f a c t that all r i g h t angles are congruent. Therefore A BAD A CAD by A . S . A . , making i B 3 Z. C. It does not take long t o see the serious error o f t h i s soc a l l e d proof. The segment as angle bisector and the perpendlcular to the base, does n o t exist -under our postulates. Moreover, the figure makes A ABC appear to be Isosceles and thus makes AD appear as introduced above. Were the f i g u r e like this,
m,
m,
-
you c e r t a i n l y would n o t consider using 755 as it is use!d . This leads us once more to say t h a t the figure 5 s merely 5 convenience
to aid you in t h i n k i n g through your reasoning in logical and c a r e f u l l y chosen words.
Problem -S e t 6-4 A, B, C and D coplanar. AD = CD. m L A =-mLC.
1. Given:
are
0
Prove; A B = C B . Does t h e proof work if A, B, C, D a r e n o t coplanar?
Qc
B
Y
2.
Given: XY = AB, AY = XB. Prove : A XOY A AOB.
3.
Given:
E, A, S and Y are
coplanar. YE S ^S.
Prove:
LY
L E
S
LA,
LS.
4. Devise a second s o l u t i o n t o Problem 5 above by introducing a u x i l i a r y segments different from the ones you used in the s o l u t i o n of Problem 3 . 5. If AC = AB and CD = BD I n the plane figure, show L A C D S L. ABD. Devise a proof that works if the figure does not lie in the plane.
A
c
B
Betweenness and Separation. Critical students may have discovered two places in Chapter 5 where the given proofs are not q u i t e complete. These d e f e c t s occur in Theorems 5-3 and 5-6, and are s i m i l a r in the two p l a c e s , consisting of a f a i l u r e t o show why a c e r t a i n p o i n t l i e s i n the i n t e r i o r of a certain angle. I n Theorem 5-3 we must know t h a t D is In the i n t e r i o r of L B A G before we can + conclude that AT> bisects t h i s angle. And I n s t e p s 9 and 10 of Theorem 5-6 we must know t h a t H is I n the interior of L ABC and of L AE^C before we can apply the Angle Addition
Postulate. I n these places I t Is not enough to observe t h a t i n the figure the p o i n t s l i e i n the proper places. Remember f i r s t t h a t a drawing is only an approximation t o t h e true geometrical situation, and secondly that this Is only one figure and t h e theorem i s supposed t o be proved f o r a l l cases. You probably wonder why an Incomplete proof should be presented In a text-book. The reason Is that the proofs of of such separation properties as t h i s one a r e o f t e n long, compl i c a t e d , and uninteresting, and that they contribute little or nothing to the essential Idea of the proof. If you understand t h e proof of these theorems as given but d i d n o t notice the incompleteness of these particular steps, you need not worry about your competence in geometry. For many c e n t u r i e s learned men d i s p u t e d whether s t e p s like these needed any justification. However, mathematicians now agree that even such "obvious" steps require a logical proof, and so we present here two theorems and some problems t o f i l l t h e gaps i n these (and l a t e r ) proofs .
Theorem 6-5. If M Is between A and C on a line L, then M and A are on the same s i d e of any o t h e r 15ne t h a t contains C.
Proof: The proof will be I n d i r e c t , If M and A are on opposite sides of L* (in the plane t h a t contains L and L') then some point D of L r lies on t h e segment 7S?I. Therefore D is between A and M, by definition o f a segment. But D lies on both L and L*. Therefore D = C. Therefore C Is between A and M. This is impossible, because M is between A and C. (see Theorem 2-3.) We can now prove a theorem which completes the proof of Theorems 5-3 and 5-6: Theorem 6-6. If M is between A and C, and B is any M point not on the l i n e AC, then M I s in the i n t e r i o r of L A B C .
Proof: By the preceding theorem, we know that M and are on the same s i d e of %?. By another application of the [sec. 6-51
A
preceding theorem (interchanging A and C) we know that M and C are on the same side of St. By definition of the Interior of an angle, these two statements t e l l us t h a t M is in the interior of L ABC, which was to be proved.
Note: figure
Problem Set On this problem s e t no Information I s to be read from a
.
Given A ABC with F between A and C, X between A and B and Q In the i n t e r i o r of A A3C. Complete the following statements, and give reasons to justify y o u r answers. a. F l i e s in the i n t e r i o r of / b. X lies In t h e I n t e r i o r of / c. Q l i e s In t h e I n t e r i o r of / L f
f
2.
The following faulty argument t h a t an obtuse angle is congruent to a r i g h t angle emphasizes t h e importance of knowing the side of a l i n e on which a lies. c
'
Suppose t h a t ABCD is a rectangle as shown and t h a t the side B5" is swung outward so that EC1 = BC and ,LABCt I s obtuse. Let the perpendicular bisector of intersect the perpendicular b i s e c t o r of BET1 at X, If X is below as shown, we have AAXD S A E X C 1 by the S.S.S. Theorem, and hence mLDAX = m L C I B X . Also, A EAX S A EBX by S . S . S . , and so m L E A X = mLEBX. It follows by s u b t r a c t i o n that mLDAE = rnLCIBE.
In case
X
lies above
<-Ã
AB, as in the figure below,
E 0 m L D A X a mLC'BX, rnLEAX = rnLEBX, and t h e desired e q u a l i t y , m LDAE = m LC^S f o l l o w s by addition. What I s wrong w i t h the above argument? A
we g e t , exactly as b e f o r e ,
c
*5.
Suppose
ABC
is a t r i a n g l e
is a point between B and C. Show t h a t if L is >L a line In the plane of A ABC < which i n t e r s e c t s a t D, then L i n t e r s e c t s "SS" or H~ ( ~ i n t : If L contains B, A 3 then L i n t e r s e c t s If L does not c o n t a i n 3 , then let Hi, Hp be the two half-planes into which L separates the the p l a n e of AAEC, H i . being the one that c o n t a i n s B. S i n c e A belongs to e i t h e r L, Hi, o r Hz, t h e r e are three cases to consider. )
and
&
m.
m.
-4.
whose truth appears obvious Is o f t e n d i f f i c u l t to prove. The following such theorem is assumed In the proof of Theorem 7-1 of t h e next chapter. Suppose ABC is a t r i a n g l e , D is a p o i n t between A theorem
+
and C and E I s a p o i n t of BC beyond C . Then each + point F of BD beyond D is in the i n t e r i o r of The t h i n g to be proved is that F is on the same side -<Ñ -<-à of BC as A and t h a t F is on the same side o f AC as E. a. How do we know that A and A D are on the same side of ¥*à BC? What theorem Implies t h a t D and F are on t h i s same side? b. Prove t h a t if Hp Kg are t h e t w o half planes I n t o <-> which AC separates t h e c plane of the figure and B belongs to H,, then each of E, F belong to "2.- This shows t h a t E and F are on the same s i d e of A C , A
LACE.
*5.
Another theorem whose truth i s f r e q u e n t l y accepted without proof is the following: If D is a point In the interior of L ABC, then intersects E. We suggest below a t r i c k y t proof I n which we consider A EAC, where E I s a p o i n t of 8 beyond B. This enables us t o apply t h e r e s u l t s of Problem 2 , Parts a and b below a r e used to show t h a t %8 does not I n t e r s e c t ES'. a . Suppose H,, Hp are the two h a l f - p l a n e s i n t o which M BC d i v i d e s the plane of
3
A EAG
In Hi. Why is D I n HA? What theorem implies t h a t each + point of BD other than B is In Hi? Why I s E i n Hp? What theorem i m p l i e s t h a t each point o f E5" other than C is in Hp? Why does f a l l to intersect BD? + Why does S? f a i l to intersect t h e ray o p p o s i t e BD? + Why does BD I n t e r s e c t 15'? + Why does the ray opposite BD f a i l to intersect A(?? with
A
-
b. c.
d.
*6.
The following theorem may be used I n s t e a d of Parts a and b of Problem 5 to show that A and C l i e on different ¥<à sides of BD. Theorem: If point D is in the I n t e r i o r of L A B C , then A is not in t h e i n t e r i o r of L DBC nor is 0 in the i n t e r i o r of L ABD. Prove t h i s theorem.
'7.
There are studies or geometry that use other systems of postulates than the ones we have adopted. A postulate taken from one such system is the following: If A, B, C, D, E are points such that A, B and C are non-colllnear and B is between A and E and D I s between B and C, then there is a point X such t h a t X is between A and C while D Is between E and X . This statement can be proved in our system of postulates. C a . Why are A, B, C , D, E coplanar? b. Show from the Plane Separation P o s t u l a t e * that ED intersects AC a t a point x be-
A and C. It can be shown t h a t
A
tween c.
0
D
is between E and X by showing that E and on opposite sides of some l i n e . What l i n e ?
8.
Given points P and Q on opposite s i d e s of plane E with Intersecting E in M. Identify the following statements as t r u e or false.
a.
b.
c.
d.
X are
^P
I
-
Lo
If L is a line in E perpendicular to PQ, then P and Q a r e on opposite sides of L in the plane determined by P and L. If L is a l i n e In E through M, then P and Q are on opposite sides of L in the plane determined by P and L, If L is a line In E, then P and Q are on opposite s i d e s of L in the plane determined by P and L. P M
-
and Q are on opposite s i d e s of every plane through n o t c o n t a i n i n g PQ.
Chapter 7
GEOMETRIC INEQUALITIES 7-1.
Making Reasonable ConJectures.
Up to now, in o u r study of the geometry of the triangle, we have been dealing only with conditions under which we c a n say t h a t two segments are of equal length, err t w o angles are of equal measure. We w i l l now proceed to study conditions under which we can say that one segment I s longer than another, (that I s , has a greater length), or one angle Is larger than another, (that I s , has
a greater measure). We shall not start, however, by proving theorems. Let us start, rather, by making some reasonable conjectureB about the sort of statements that ought to be true. ( ~ h e s estatements should not be called theorems unless and until they are proved.) An example: Given a triangle with two aides of unequal length, what can we aay about the angles opposite these s i d e s ? Notice that t h i s problem is naturally suggested by Theorem 5-2, which says that if t w o sides of a triangle have the same length, then the angles opposite them. have the same measure. You c a n Investigate t h i s situation by sketching a triangle with two a i d e s of obviously unequal lengths, like t h i s :
Here BC is greater than AB, and m L A I s greater than m/- C. After sketching a f e w more triangles, you will become pretty w e l l convinced that the following statement ought to be true: If two aides of a triangle aFe of unequal length, then the angles opposlte t h e m are unequal measure, -and the larger angle
-----
LEI oppo8ite the longer stde. -
Now try the same sort of procedure with the following problems.
Problem Set Here are some experiments for you to try. 1. Consider triangles w i t h two angles of unequal measure. Write a statement which you think may be true concerning the sides opposite those angles. 2. Consider several t r i a n g l e s ABC. How does AB + X compare with AC? EC + AC compare w i t h a? A B + AC compare with EC? These responses suggest a general conclusion. If you think t h i s conclusion is true f o r a l l triangles, write it as a proposition. 3. Consider a quadrilateral RSTQ. How does RS + ST + TQ compare w i t h RQ? Write a proposition suggested by your answer. 4. Draw several triangles In which the measure of one angle I s successively greater but the adjacent s i d e s remain unchanged in l e n g t h . What happens to the length of the t h i r d s i d e ? 5. Draw A DEF and A XY2 such that DE = XY, FS = ZY, and m L DEF > m L XYZ. Compare DF and XZ. 6 . Regarding A PDQ and A JUN such that m JU, and QD = NU, a hasty person might conclude t h a t PQ > JN. Draw a figure showing t h a t the conclusion is not
.
just ifi e d 7 . A is a point In plane E, TS' is a ray n o t lying in E, and * AC is a ray lying in E. Con-
L *
sidering d i f f e r e n t positions + of AC, describe as accurate! -3 as you c a n t h e position of AC which makes L BAG as small as p o s s i b l e ; as large as possible. .No proof I s expected but you are asked to guess the answer on t h e basis of your knowledge of space.
On the basis of drawings decide whether or not an angle can be trisected by the following procedure: Let A ABC be an i s o s c e l e s triangle w i t h congruent sides A 3 and ^(T. Trisect s i d e with p o i n t s D, E so that BD = DE = EC. Is f BAD Z L DAE 2 i EAC?
8.
-
B
0
E
C
7-2. Algebra
Inequalities. Before considering geometric Inequalities we review some of t h e f a c t s concerning i n e q u a l i t i e s between real numbers. Note f i r s t that a < b and b > a are merely two ways of writing the same t h i n g ; we u s e whichever is more convenient, e . g . 3 < 5 or 5 > 3. Definitions. A real number is positive If It is greater than zero; It is negative If I t I s less t h a n zero. We now r e s t a t e t h e o r d e r postulates, giving examples of t h e i r
use.
0-1. (uniqueness of order.) For every x and y, one and only one of t h e following r e l a t i o n s holds: x < y, x = y, x > y. 0-2. (~ransitivityof o r d e r . ) If x < y and y < z , then x
<
2.
Example 1. Example 2.
Example 3. Given; To prove:
hence, 3 < 9. If we know t h a t a < 3 and b > 3, we can conclude that a < b. Proof: If a < 3 and 3 < b, then a < b. Any positive number is greater than any negative number. p I s positive, n is negative. p > n. 3
<
5 and
Proof : 1. p is positive. 2. p > 0.
5
< 9,
1. 2.
Given. Definition of posltlve.
Relation between < and Def l n i ti on o f negative Postulate 0-2. Relation between < and
3. 4.
.
5. 6.
x
+
>. >.
0-3. (Addition for i n e q u a l i t i e s . ) If x < y, then z < y + 2, for every z. Example 4. Since 3 < 5 I t follows t h a t 3 + 2 < 5 + 2, or 5 < 7; that 3 + ( - 3 ) < 5 + ( - 3 ) , or 0 < 2; t h a t 3 + (-8)< 5 + ( - 8 ) , or -5 < - 3 . Example 5. If a < b then -b < -a. Proof: a + (-a-b)< b + ( - a - b ) , or -b < - a . Example 6 . If a + b = c and b is positive, then
a
<
c.
Proof: 1. b I s positive. 2. b > 0. 3. O < b . 4. a < a + b .
5.
Example 8.
Why? Why? Why?
4.
Why?
5. MY? then a < c
a < c .
Example 7 .
1. 2. 3.
If a + b < c - b. Proof left to the student. If a < b, then c - a > c - b for every c. Proof l e f t to the student.
<
z
>
0-4. (~ultiplicationfor ~ n e q u a l i t i e s . ) If x y and 0, then x z < yz. Example 9. From 3 < 6 we can conclude that 3000 < 1
6000; also, that -g- - 3 < & - 6 , o r
and z < 0 , then xz Proof l e f t to the student.
Example 10, If x
5 < 71 .
> yz,
(Addition of ~nequalities.) If a < b and x < y, then a + x < b + y . This Is not a postulate but a theorem; I t s proof is given in Section 2-2. However, I t is convenient to list it, f o r
0-5.
reference, along with the postulates.
7-3.
-The Basic
Inequality Theorems. In the figure below, t h e angle L E D is c a l l e d an exterior angle of A A X . More precisely:
If C is between A and D, t h e n L BCD Is an exterior angle of A ABC. E v e r y triangle has s i x e x t e r i o r angles, as I n d i c a t e d by the double -headed arrows in the figure below: Definition.
These six angles form three pairs of congruent angles, because they form three pairs of vertical angles. Definition. L A and LB of the triangles are called the remote Interior a n ~ l e sof the exterior angles L E D and L A C E . Similarly, L A and L C of A A N are the remote I n t e r i o r angles of the exterior angles L ABF and LC%. Theorem 7-1. h he Exterior Angle Theorem. ) An exterior angle of a triangle is larger t h a n e i t h e r remote i n t e r i o r angle. Restatement: N o t e first that the two exterior angles at vertex C, above, have equal measures ( v e r t i c a l angles), and so it
doesn't matter which of them we compare with L A a n d L 3 . It turns out to be easiest to compare A L BCD with mL B and m L ACE w i t h mL A . Since t h e proofs of these two c a s e s are exactly similar we need prove o n l y one. Given triangle ABC. If C is between A and D, then mLBCD
> m L
B.
Proof: Reasons
Statements 1.
Let
E be the mid-point of E. 1.
Ety Theorem 2-5 t h e r e is such a mid-point.
2.
Let
F be a o nt of the ray
2.
By Theorem 2-4, there is such a p o i n t .
3.
Vertical angles are eongruent . Statements 1, 2, 3 and t h e S.A.S. Postulate, Corresponding parts of congruent triangles. Postulate 13 (The Angle Addit i o n Postulate. ) Statements 5 and 6. By algebra from s t e p 7. (Since rn L FCD is a p o s i t i v e number, Example 6 o f Section 7-2 applies.)
opposite to
EF
=
EA.
@ such t h a t
4.
5. 6.
m L B C D = m L E O F + m L FCD.
6.
7. 8.
Problem Set 7-3a 1.
a.
Name the remote I n t e r i o r angles of the e x t e r i o r angle L ABE in t h e f i g u r e .
b
2.
.
a.
and L BAC are the remote i n t e r i o r angles of which exterior angle? LA X
E
In the figure, which angles are e x t e r i o r angles of the triangle?
b.
What I s the relationship of m L DBC fco m L A? Why?
c.
What is the relationship of
m d.
3.
L
DBC
to
m L C?
Why?
What Is the r e l a t i o n s h i p of m L DBC to m L CBA? Why?
U s i n g t h e f i g u r e , complete the following: a. If x = 40 and y = 3 0 , then m L BCE b . If x = 72 and y = 73, then m L BCE c.
If y = 54 and m L BCE
d.
If m f BCE = 112, then x
e.
If m L BCE = 150, then z
f.
If x
=
g
If
z = 68, then
25 and
m L BCE
x = 90 and
>
2 =
90, then
a
y = 90, then
A
C
E
196 4.
The accompanying figure I s an illustration o f t h i s statement: An exterior angle of a quadrllateral is greater than each of the remote I n t e r i o r angles. Is t h i s a true statement? Explain.
*5. Prove the following theorem: The sum of the measures of any t w o angles of a triangle i s l e s s than 180.
Given:
A ABC with angle
Prove:
measures as in the figure a + b < 180. b + c < 180. a + c < 180.
B
A
E
\
.
* 6 . Prove the following theorem: The base angles of an i s o s c e l e s triangle are acute. (~int; Base your proof on t h e statement of the p r e v i o u s problem. )
Theorem 7-1, w h i l e perhaps not very exciting in Itself, is extremely useful in proving other theorems. ( A theorem of this type is sometimes c a l l e d a lemma.) For* example, the following is a u s e f u l corollary. Corollary 7-1-1. If a triangle has a right angle, then the o t h e r two angles are a c u t e .
Proof: If m L A = 90, then m ^. BCD > 90, and therefo r e , m L EEA < 90. In a similar way we can prove m LAX < 90. We next use Theorem 7-1 t o prove two more congruence theorems. Theorem 7-2. (The S . A . A . heo or ern) Given a correspondence between two triangles. If two angles and a side opposite one of them in one triangle are congruent to the corresponding parts of t h e second t r i a n g l e , then t h e correspondence is a congruence. Restatement; L e t ABC-DEF be a correspondence between two t r i a n g l e s . If L A ~ D, L L B ~ L E , 15" Z'BF, and A ABC = A DEF. then Aril
Proof:
1.
2.
+
A 3 take AX = DE.
On
Reasons
Statements
X
A A X C ~ ADM.
so that
1. E .
Point P l o t t i n g Theorem. S.A.S.
~ostulate. 3 . m L A X C = m L DEF. 3 . D e f init Ion of congruence. 4. m L A X C = m L A X . 4 . Step 3 and given. Now suppose t h a t X is not t h e same point a s B.
5.
Either X Is between A and B or B is between A and X.
5. Step
1 and definition of ray.
6. In either case one of L A X C
6. Definition of exterior angle
7.
m
8.
X=B.
7. 8. 9.
and L ABC is an e x t e r i o r angle of ABXC and t h e o t h e r I s a remote I n t e r i o r angle.
9.
LAXC
+mL
ABC.
A A B G Z A DEF.
and remote I n t e r i o r angle.
Step 6 and Theorem 7-1. S t e p 7 contradicts Step 4. Steps 2 and 8.
Although I t was pointed o u t in connection w i t h t h e S.A.S. Postulate t h a t an S.S.A. theorem cannot I n general be proved, there is one special case; namely, the case in which the angle is a right angle, that follows from Theorem 7-2.
h he
Hypo tenuse-Leg he or em. ) Given a correspondence between two right triangles. If the hypotenuse and one leg of one triangle are congruent to the corresponding parts of the second triangle, then t h e correspondence Is a congruence.
Theorem 7-3.
Let
Restatement: In AABC andA DEF l e t m L A = m L D = 90. ABC-DEF' be a correspondence such t h a t
Then
A ABC 2 A DEF.
DQ = AC.
EQ = BC. In A DB&
-
take Q such that Then & DEQ ZA ABC by the S.A.S. Postulate, and so A EQF I s thus an I s o s c e l e s triangle, and SOL EQD 2 EFD. and A DEF we thus have
Proof;
On the ray opposite to DF
.
Theorem, A DEF 2 A DEQ Since we have already e s t a b l i s h e d A DEQ 2 A ABC we conclude t h a t A DEF 2 A A X , which I s what we wanted.
Hence, by the
S .A. A.
Problem S e t 7-3b 1.
If in this figure
2.
Given that
AQ
=
BQ
mlm, AB = HF, AK =
Prove that
3.
If
AX =
FH
prove t h a t
and
KF = QB.
HQ.
,,
In this figure, FB = AB.
it.
If two altitudes of a triangle are congruent, the triangle is isosceles.
5.
In this figure: L C 2 L a . AQ = AF.
Prove: OS = FK.
6.
B
In this figure i f L a 2 L c,
-ABJAH
and
prove that
Elm, AH
=
F
FH.
*
If two s i d e s of a triangle are not congruent, Theorem 7-4. then the angles opposite these two s i d e s are not congruent, and the larger angle is opposite the longer s i d e . Restatement:
B
Given A ABC . If AB
-- ------
\
> AC ,
then m L C
> mL B.
\C\\
--- --
\
- \
\
\
\
--lD
Proof: Let D be a p o i n t of 2, such that AD = AB. ( ~ the y P o i n t Plotting Theorem, there is such a point. ) Since the base angles of an I s o s c e l e s triangle are congruent, we have
(11 rn f ABD = mL D. Now AD > AC, slnce AD = AB and A 3 > AC, and so C I s between A and D by Theorem 2-1. Theorem 6-6, C is in t h e i n t e r i o r of ABD, and so m f ABD = m L A E 4- rnL CBD (2) by t h e Angle Additlon Postulate. S1nc-e m L CBD > 0 it follows that
m L ABD
(31
>m
L
A=,
Therefore
(41 Slnce ACB
w
rnL D > m L AX, from (I) and (3). i8 an e x t e r i o r angle of A X D , we have m L ACB > rn L D.
(5) ( 4 1 and ( 5 L
mf. ACB
>
th L A=,
that I s ,
m L c > m L B, which was t o be proved.
T h e o ~ e m7-5. If two angles of a triangle are not congruent, then the sides oppo3ite them are n o t congruent, and t h e longer s i d e Is opposite the lapger angle.
Restatement:
In any triangle A A X , Tf m L C
>m
L B, then
A 3 3 AC.
We want t o prove t h a t AB > AC. S m c e AB and AC are numbers, t h e r e are only three possibilities: (1) AB = AC, (2) AB < AC and ( 3 ) A 3 > AC. The method of the proof is to show that the f i r s t two of these ~ ~ p o s s l ~ i l ~ are t l e ~in' lfact impossible. The only remaining p o s s l . b i l i t y will be (31, and t h l s will mean Proof:
t h a t the theorem is t r u e . (1) If AB = AC, then by Theorem 5-2 it follows that L B 2 L C; and this is false. Therefope, I t is imposa3ble that A3 =
AC.
<
AC, then by Theopem 7-4 I t follows thaC m L B; and t h i s is f a l s e . Therefore, It is impossible
(2)
If
AB
m L C < t h a t A% < AC. The only rernainlng pussibLllty is t h a t A 3 > AC, whlch was to be proved. The proof of Theorem 7-5, a s we have given it, I s r n e ~ e l ya handy way of stating an l n d i ~ e c tproof. It could have been written mope f o r m a l l y y llke this; "Suppose that Theorem 7-5 is false. !hen e i t h e r AB = AC or AB < AC. It is impossible t h a t A 3 = AC, because And I t I s irnpossTble that AB < AC, because Therefore, 7-5 I s not false The proof is probably easier t o read, however, the way we gave it the first t i m e . We will be using t h e same sort of scheme again. That is, we will list the possibillt~es, in a given aituation, and then show that all but one of these ''posslbi1lt~es~~ are in f a c t ImpossEible; it will then f o l l o w t h a t the last remaining p o s s i b l l ity must represent what actually happens. 11 T h a t process starts upon t h e supposition t h a t when you have eliminated all which is Impossible, then whatever remains, how eve^ i m p ~ o b a b l e must be t h e t r u t h . " (sherlock Holmes in " T h e Adventurn of the Blanched ~ o l d l e ~ " . Theorems 7-4 and 7-5 are related in a s p c i a l way; they are called converses of one another. To g e t one f ~ o mthe other, we interchange the hypothesis and the conclus5on. Me can exhib3t thls f a c t by restating t h e theorems t h i s way: Theomm 7-4. Given A A=. If A 3 > AC, then m L C > m L B. Theorem 7 - 5 . Given A A=. I f m L C > m L B, then AB > AC. We have seen lots of p a l m of theorems t h a t are related t h i s way. For example, we showed t h a t if a trlangle is isosceles, then I t s base angles are congment; and l a t e r we showed that If t h e base angles of a t~iangleare congruent, then the triangle is
."
....
....
isosceles. Each o f these theorems is the converse of t h e other. We showed that every equilateral triangle 1s equiangular; and l a t e r we proved the c m v e r s e , which s t a t e s t h a t every equiangular t ~ i a n g l e2 s equilateral. It is v w y important to remember that the converse of a true theorem is not necessarily t r u e a t all. For example, the theorem 1'vert5cal angles are congruent" is always true, but the converse, "congruent angles are vertlcalt' 1s c e r t a i n l y not true In all cases. If two triangles a r e congruent, then tney have the same area, but if t w o triangles have the same area, it does not follow that they are congruent. If x = y, then it follows t h a t x2 = y2 ; b u t I f 2 2 x = y , it does n o t f o l l o w that x = y. h he o t h e r p o s s i b i l i t y 1s that x = - y . ) It is true t h a t every physicist 1s a s c i e n t f s t , but it is n o t t r u e t h a t e v e r y scientist is a physicist. If a theorem and its converse a r e both true, they can be convenlentl$ combined into a single statenent by w i n g the phrase "if and only i f 1 ' , Thus, if we say: Two angles of a t r i a n g l e are congruent if and only if t h e oppo s ft e s i d e s a r e congruent ; we are including 3n one statement both theorems on isosceles tr3angles. The first half of t h i s double statement: TWO angles of a t r i a n g l e are congment If t h e o p p o s i t e sides are congruent; is Theorem 5-2; and t h e second h a l f : Two angles of a triangle a p e congruent only if the opposlte s i d e s are congruent; is a restatement of Theorem 5-5*
P ~ ~ b l m 7-3~
1.
2.
In AGHK, GH = 5, HK = 14, KG = 11. Name the largest angle. Name the smallest angle. In A AEC, m L A = 36, m L B = 74, and m L C = 70. Name t h e largest side. Nme t h e shortest s i d e .
3.
Given the figure with HA = HB, m L HEX = 140, and m L A m = 100, fill in the b l a n k s below: b.
m L A = m L RHB
c.
---
a.
---. =
---
I s t h e longest s i d e
A
B
K
of A ABH.
4,
What conclusion can you reach a b u t the length of A KLM if: a. m L K > m L M? b. m L K < m L L ? c . r n L M > m L K > m L L? d.
in
rnLM>mLL?
m L K > m L M and m L K > m f L? f. m L K > m L L and m L M < m L L?
e.
5.
If t h e f l g u r e were c o r r e c t l y dram which segment would be the longest?
c
. 6.
Name the sides of the figure In order of increasing lewh.
c 7.
-
If in the figure AF I s the s h o r t e s t s i d e and is the longest s i d e , prove t h a t m L F > m L B. (Hint! w e diagonal FB.) 4
*8. If the base of an isosceles triangle is extended, a segment which joins the vertex of the trtangle w i t h any point In this extenszon is greater than one of the congruent sides of the triangle. F
9.
Try to d e c i d e whether each statement, and each converse, is true or false. a. If a team has some spirit, it can win some games. b. If two angles are right angles, they are congruent. c. Any two congruent angles are supplementary. d . The i n t e r i o r of an angle is the i n t e r s e c t i o n of two Write the converse of each statement.
half-planes. e . If Joe has scarlet f e v e ~ ,he is s e r i o u s l y ill. f. If a man lives in Cleveland, Ohio, he lives in Ohlo. g. lf the tmee angles of one triangle are congruent to the corresponding angles of another triangle, the triangles are congruent. h. If two angles are complementary, the sum of their measures I s go. 10. When asked to give t h e converse of t h i s statement, "If I hold a lighted match too long, 1 will be burnedn, John said, ''1 w311 be burned if 1 hold a IZghted match t o o long." Was Johnts sentence the converse of t h e o r l g z n a l statement? D~SCUSS 11. a. Is a converse of a true statement always t r u e ? Which parts of Problem 9 illustrate your answer? b. May a converse of a false statement be t r u e ? Which parts
.
o f Problem 9 Illustrate your answer? [see.
7-31
Theorem 7-6. The shortest segment Joining a point to a line is the perpendicular segment.
Restatement: L e t Q be t h e foot of the perpendicular to the line L through the point P, and l e t R be any other point on L. Then PQ < PR. Proof: Let S be a point of L, such that Q is between S and R . Then L PQS I s an e x t e r i o r angle o f A PQR. Therefore, m L PQS > m L PRQ. But mL PQS = m L PQR = 90, and so m L PQR > m L PRQ. By Theorem 7-5 it follows that PQ < PR, which was to be proved.
---
The distance between a line and a p o l n t not on it is the length of the perpendicular segment f r o m t h e point t o the l i n e . The distance between a l i n e and a point on the line l a def i n e d to be zero. &finition,
Theorem 7-7. (The Triangle Inequality.) The sun of the lengths of any t w o sides of a triangle Is greater t h a n t h e length of the third s i d e . Restatement: In any triangle A A X , we have AB + BC > AC.
P r o o f : Let D be a point of the ray opposite to that Dl3 = AB. Since B is between C and D
+
DC=DB+BC. (1) DC = A B + BC.' (2) m L DAB < rn L DAC,
Then Also
B is in the i n t e r i o r of L DAC. Since A DAB is i s o s c e l e s , w i t h AD = DB, it follows that because
( 3 ) m L AD3
= m /.
DAB.
By ( 2 ) and (3) we have m L ADB Applying Theorem
7-5 to A
ADC,
DAC.
we see- t h a t
( 4 ) DC
>
AC.
By (1) and ( 4 ) it follows t h a t
AB + BC
> AC,
which was t o be proved.
Problem Set 7'-3d 1.
Here
AH
m<-
< -and and BT <
the theorem invo 1ved
AH
.
< .
. State
I A
EC
such
2.
With angle measures as shown
in the f i g u r e , I n s e r t HA, HF, HB below In correct order.
State theorems t o support your conclusion.
3.
4.
Suppose that you wish to draw a triangle with 5 as t h e length of one s i d e and 8 as the length of a second side. Your third s i d e must have a length greater than -' andlessthan
Suppose t h a t you w i s h to draw a triangle w i t h j a s t h e length of one s i d e and k as the length of a second side. It is known t h a t j < k. Indicate, as e f f i c i e n t l y as you can, the restrictions on the length, x, of the t h i r d s i d e .
Prove t h a t the sum of the lengths of the diagonals of this quadrll a t e r a l I s less than the sum of the lengths of its sides. Given: Quadrilateral ABCD. To prove: DB + CA < A 3 + BC
*6.
.
+
CD
+
A , B, C, be p o i n t s , not necessarily d i f f e r e n t . Prove t h a t AB + BC > AC and t h a t A 3 + BC = AC if and only if B I s on t h e segment T^.
Let
*7.
*8.
Prove that the shortest polygonal path f r o m one point to another Is the segment Joining them.
Given:
n
Prove:
A A
points
1 2
Al,
An,
......,
+ AgAg+ ..... +
Given two segments
and
*n
2 AIAnIntersecting at
P.
Prove that If X is any point in the plane of ABCD o t h e r than P, then XA + XB + XC + XD > PA + PB + PC + PD. Will this result be true if X is n o t in the plane of ABCD?
*9.
Given a line m Find the point
possible.
and two p o i n t s P, Q on the same s i d e of m. R on m f o r which PR + RQ I s as small as
We w i l l now prove a theorem which is a l i t t l e like Theorem 7-5, except t h a t it deals w i t h t w o triangles instead of one. Theorem 7-8. If two sides of one triangle are congruent respectively t o two sides o f a second triangle, and t h e included angle of t h e f i r s t triangle is larger t h a n the Included angle of the second, then the o p p o s i t e side of the f i r s t triangle is longer than t h e opposite side of the second. Restatement: Given A ABC and& DEF. If AB = DE, AC = DP and r n f A > rn L D, then BC > EF,
-
B
Proof: Step 1. We construct A AKC, with o f L BAG, such t h a t A AKC 2 A DEF, like t h i s :
K
in t h e i n t e r i o r
To do t h i s , we use t h e Angle Construction P o s t u l a t e , t o get a ray ¥<-à w i t h Q on the same side of AC as B such that L QAC 2 L D. On AQ^ we take a p o i n t K such t h a t AK = DE. 'By t h e S.A.S. P o s t u l a t e , we now have A AKC 2 A DEF, which is what we wanted.
s,
Step 2.
Now we b i s e c t LBAK, and let
t h e b i s e c t o r crosses
E,like
14
be t h e p o i n t where
this:
The marks on t h e f i g u r e indicate that AK = AB, and t h i s is true, because AK = DE and DE = AB. We a r e now almost done. Ey the S . A . S . Postulate, we have A ABM 2 A AKM. Therefore, MB = MK. By Theorem 7-7, we know t h a t
CK
<
CK
< OH +
CM + MK.
Theref ore,
because MB = MK. Since CK = EF EF < BC, which is what we wanted.
MB,
and
CM + MB = BC, we g e t
The converse of thie theorem 1s also true, Theorem 7-9. If two sides of one triangle are congruent respectively to two s i d e s of a second triangle, and t h e third s i d e of t h e f i r s t triangle is longer than the t h i r d side of the second, then the included angle of the f i r s t triangle is larger t h a n t h e included angle o f the second. The proof Is similar to t h a t of Theorem 7-5, use being made of Theorem 7-6 and t h e S . S . S . Theorem t o eliminate the t w o unwanted cases. The student should fill In the details.
-
Problem Set 7 - 3 e 1.
S t a t e the combination of Theorems 7-8 and 7-9 In the
"If and
only if" form. 8
2,
In this figure AC = BC, and BD < AD. Prove: mL. x > m L y.
3.
In i s o s c e l e s triangle RAP with
-
and B a p o i n t on AF such that m L ARB < m L BRF. Prove: A 3 < BF. RA = RF
4.
Given A ABF with median and m L ARB = 8 0 . P r o v e : m L A > m L F.
5.
In A A B C , BC
>
AC
obtuse or a c u t e ?
and
Q is the midpoint of Explain. [ s e c . 7-31
-
AB.
Is /.
CQA
6.
In t h i s f i g u r e AH > F'Q. Prove: AB
FH =
AQ.
> FB.
A non-equllateral quadrilateral has two pairs of congruent adjacent s i d e s . Prove t h a t the measure of the angle included between the smaller s i d e s Is greater t h a n the measure of the angle between t h e larger sides.
Prove the following theorem: If a median of a triangle is not perpendicular* to the s i d e to which it is drawn, then t h e lengths of the other two sides of t h e triangle are unequal.
9
Given AB 3 AC and FC = DB in t h i s figure. Prove that
FB
> CD.
A
8
c
F
214
7-4.
Altitudes.
Defini,tion. A n altitude of a triangle is the perpendicular segment Joining a v e r t e x of the triangle to the line that contains the opposite s i d e .
--
-
e I s called the altitude from B to AC, or simply the altitude- from ( ~ o t t c ethat we say the a l t i t u d e from B instead o f an altitude from B, because Theorem 6-3 t e l l s us that there I s only one. ) Notice t h a t the f o o t of the perpendicular does not necessarily lie on the s i d e AC of the triangle. The figure may look l i k e
In the figure
this:
-
Notice also that every triangle has three altitudes, one from each of t h e three v e r t i c e s , like t h i s :
Here T^F I s t h e altitude f r o m A , and is t h e a l t i t u d e from C.
BD
I s t h e a l t i t u d e from
B
It is customary to use the same word " a l t i t u d e " f o r t w o o t h e r d i f f e r e n t , but related, concepts. (l) The number which is the length of t h e perpendicular segment Is called altitude; thus one may say h he a l t i t u d e from B is 6 " , meaning that BD = 6. (2) The line containing the p e r p e n d i c u l a r segment is a l s o c a l l e d a l t i t u d e ; a p r o p e r t y of the above f i g u r e c a n be expressed by saying that the three altitudes of the triangle Intersect In one point. (This property is true f o r a l l triangles and will be proved in Chapter 14. ) This t r i p l e use of the one word could cause t r o u b l e but generally does not, since it is usually easy to tell In any particular case which usage is being made.
--
Problem Set 7-4 1.
2.
Define:
Altitude of a triangle. Median of a triangle. Draw an obtuse triangle (a triangle having an obtuse angle) and its three altitudes. a. b.
[sec.
7-41
In an equilateral triangle a median and an altitude are drawn to the same s i d e , Compare the lengths of these two segments.
!
Prove that t h e perimeter o f a t ~ l m g l eis greater than the sum of the three a l t i t u d e s .
1
Prove the following theorem: triangle are congruent,
The altitudes of
I
I
an equilateral
Review Problems
Three guy wires of equal length are being used t o support a newly planted tree on level ground. If they are a l l fastened t o the t r e e at the same height on the tree, will they be pegged t o t h e ground at equal d l s t ~ c e sfrorn %he foot of the
tree? Why? If t h i s figure were drawn correctly, which segment in the figure would b e $he ~ h o r t e s t ? Explain your reasoning.
A
c Prove the following theorem: If two oblique ( n o t perpendicular) line segments are d~awnt o a l i n e from a p o l n t on a perpendicular to t h a t line, the m e containing the polnt more remote frorn the foot of the perpendicular is t h e longer.
In this planar figme, AK = HQ, AF = HB, Prove f Q g L !A. Bes bisect BF?
LK
WlE.
AC > AB. Prove EC between B
5.
In & A E , point on
6,
Segments drawn from a point in the Interior of a t r t a n g l e t o t h e three v e r t i c e s have lengths r* s, t. Prove that r +- s + t I s g ~ e a t e rthan half' the pertmeter of the triangle.
7,
I n t h i s planar f i g w e the shor%est s i d e and t h e longest side. Prove
mL F
8.
*g.
-
t h a t m y l i n e segment from A and C 1 s s h o r t e r t h a n AC.
is is
to a
H
>mLA.
Prove t h e following theorem: The length of the longest side of the t r i a n g l e I s less than half its perimeter,
Given Isosceles A Am w i t h FA = l?B, AB
<
H on AF, so t h a t F is between A and H, Prove no two s i d e s of A ABH are e q u a l in l e n g t h .
H
AF, and
A
*lo. On the basis of t h e a s ~ m p t l o n swe have accepted and the theorems we have proved in this course we are not able at present to prove that t h e sum of t h e measures o f t h e t h r e e angles of a t r i a n g l e is 180 (an idea w l t h which you have been familiar for some time). But* we can e a s i l y construct a trlangle and prove that the sum of the measures of the angles of this triangle is less than 181. Let L FCG have measure 1 { Angle A c Construction P o s t u l a t e ) . On CF I*
-
+
-
take points A and 3 8 G so t h a t CA = CB ( P o ~ n tPlotting heo or em), Why i s t h e sum of t h e measures of the angles of this t r i a n g l e less than 181? and
CG
*ll,
Tine sum of t h e measures of the three angles of a triangle is less than 270.
*12. In this f l g u r e : L C 1s a right angle. m L 3 = 2 m L A.
Prove: AB = 2 CE. (Hlnt: Introduce auxlliaxy segments,)
Prove this theorem: The sum of the distances from a point w l t h i n a triangle to the ends o f one ~ i d eis less thm t h e sum of t h e lengths o f t h e o t h e r t w o s i d e s , w *14, Su.ppose AC i n t e r s e c t s 3D at a p o i n t B between A and C , w Pe~pendicularsare dropped from A and C to BD BtPlkIng it at P and Q ~ e s p e c t i v e l y . S~IOW that P and Q are not on the same side of 33.
*l3.
*
Chapter 8 PERPENDICULAR LINES AND PLANES IN SPACE
8-1.
The B B I CDeflnitlon, In t h l s chapter we s h a l l be specifically concerned with properties of figures t h a t do not lie in a single plane. The fundamental properties of such figures are s t a t e d in Postulates 5b, 6, 7, 8 and 10, and inTheorems 3-2, 3-3 and 3-4. It would be worth your wh5le to review these. Definition. A l i n e and a plane are perpendicular If they I n t e r s e c t and if every llne lylng in the plane and passlng through the polnt of intersection is perpendimlap to t h e given line.
ELL.
If l b e L and plane E a r e perpendicular we write L ~ Eor W e have indicated, i n the figure, three lines in E passing through P. Notice that in a perspecttve drawtng, perpendicular llnes dontt n ~ c e s s a r ~ llyo o k perpendicula~~ Notice a l s o t h a t if we merely required t h a t E c o n t a i n one l i n e through P p e r p e n d i c u l a ~ t o L, t h i s would m e a n very l i t t l e ; you can fairly e a s i l y convlnce you~selfthat every plane through P contalns such a line.
Problem -Set 8-1 1.
The f l g u ~ eat t h e r i g h t represents plane E. Do any points outside t h e qua& i l a t e r a l shown belong t o plane E ? Is plane E Intended to i n c l u d e e v e r y p o l n t outside t h e quadrilateral?
/ u
Sketch a plane perpendicular t o a vertical line. (See Appendix V, 1 Sketch a plane perpendicular t o a h o r i z o n t a l llne. Does each of your sketches represent a llne perpendicular to a plane?
Repeat the sketch of Problem 2b. Add to t h e sketch t h r e e l h e s In the plane which pass through t h e point of i n t e r section. What I s the relatLonship between each of the three lines and the orfginal line? 4.
Reread the deflnitlon of perpendicularity between a llne and a plane and decide whether the following statement is t r u e if' that definition i s accepted: TI If a l i n e is perpendicular to a plane, then it is perpend i c u l a r t o every line lylng In the plane and passing through the p o i n t of intersection. "
5*
that B, R, S and T a r e I n plane E, and t h a t AB E, whlch of t h e fol~owlngangles must be r l g h t angles: L Am, Am, L R B T , L TBA, Given
-
1
L SBR?
\ !--jj/ T
E
I
4
6.
If L PQH is a r i g h t angle and Q and H are jn E, should you infer from t h e definition of a line and e a plane perpendicular t h a t PQ E? Why o r why not?
1
7.
In the figure plane E contains points R, S, and P, but not T. a, Do points R, S and T determine b,
a plane? + If SP is perpendtcular to the plane of R, S and T, which angles In the figure must be r i g h t angles?
8.
a. b.
*g,
a.
b.
10.
M
If a p o i n t 5s equldlstant from each of two o t h w p o h t s , are t h e three polnts c o p l a n a r ? If two p o i n t s are each equidistant from each of t w o other p o h t s , are the f o u r p o i n t s coplamar? Given: Collinear points A, B and X as in the figure; I3 equidistant from P and Q; and A equldlstant f r o m T and Q. Prove: X is equidistant from F and Q. b e s the proof require t h a t Q be in the plane o f A, B, X and P?
P
Q
Look ahead t o Theo~em8-1 and make a model for it from s t i c k s , wire c o a t hangem, or straws.
8-2.
The Easic Theorem. The b a s k theorem on perpendicularity In space says t h a t if a plane E c o n t a i n s t w o llnes, each perpendicular to a line L at the we same point o f L, then L E. The proof of t h i s Is easler If prove two preliminary theorems (lemmas} Theorem 8-1. Xf each of two points of a 13ne I s equldlstant from t w o given polnts, then every point of the line is equidistant
1
.
Restatement: If P and Q are two p o i n t s and L is a l i n e such t h a t two p o i n t s A, B of L a r e each equidistant from P and Q, then he above figure every point X of L is equidistant from P and Q. 8hows t h r e e p o s s i b l e positions f o r X. Of c o u r s e , X might be a t A or B,) Proof: First we c o n s i d e r t h e case whem X I s on the sane s i d e of A as B. X mLght be a t X B, or X but for convenience 1' 2 in the flgure we show it beyond 3 at XI. In this case LPAB =LPAX and L QAB = L Q h X , We t r e a t t h i s case in 3 steps.
1.
Since AP
=
AABP Z f i
(given), and AB = AB (identity), BP = Hence, L PA3 Z L QA3. (S.S.S.).
AQ ( g i v e n ) ,
Am
2.
,L QAX. This is because L PA3 2 L QAB by Step 1. (we are considering the c a s e where L PAX = L PAB and LQAX = L WB.1
3.
Using Step 2 and t h e f a c t s t h a t AP = AQ ( g l v e n ) , and AX = AX ( i d e n t i t y ) we flnd that A PAX 2 A W (s,A.s). Hence PX = QX.
L PAX
+
The case where X ltes on the ray opposite AB 1s proved in a similar f a s h i o n .
1.
A piece of paper AXE&
as pictured here, 1s f o l d e d along Imagine A and B as both being in the foreground of the picture and in the background. Under these conditions will a polnt K of be equtdistant from A and B? State a theorem to support your answer. If AF = 6, 3F = *
m*
2. Here mine plane AXE obscuring part of plane A m . It I s given that XA = X% and YA = YB, T, W and 2 ape three other p o i n t s of XY. b e 8 TA = TB?
-
b e 8 WA = WB? Lbes ZA = ZB? State a theorem that supports each conclusion.
6
. \
If each of three non-collinea~p o i n t s of a plane is e q u i d i s t a n t from t w o p o l n t s , then every p o l n t of' the Theorem 8-2.
p l a n e 2s equidistant from these t w o points, Given: T h e e non-colllnear polnts A, B and C each equidistant from P and Q ,
0
P
Prove: Zvery p o i n t of t h e plane determined by A, B ahd C Is equldlsta,nt from P and Q The proof is given in t h ~ e esteps.
Proof:
.P
1.
Since A and B are each glven equidistant from P and Q, each M p o i n t of AB is equidistant from P and Q. This follows f ~ om / Theorem 8-1. Similarly each * p o l n t of I X is equidistant from
/// ..
/ Q
P and Q. 2.
L e t X be any other p o l n t of the H plane. If X I s on e i t h e r AB or ++ CB, X i s equidistant from P and Q by S t e p l . If X is on one side of W , choose Y, some p o t n t w of AB ori the other side of ?%.
-
-
The Plane Separation Postulate assures us that there Is such a p o i n t Y and t h a t XY will I n t e r @ s e c t C B in some polnt 2. 3*
aP
/ A T Q P
w
Slnce Z is on CB it I s equ3distant from P a Q by Step 1. since Y i s m ttisequid i s t a n t from P and Q by S t e p 1 Therefore by Theorem 8-1 every p o l n t of YZ is e q u l d 3 s t a n t from P and Q. points. [ sec. 8-21
%
X
@Q is one of' these
Since we have shown t h a t each point X of the plane determined by A, B, C is equidistant from P and Q, Theorem 8-2 I s established. We are now ready to prove the basic theorem.
Theorem 8-3. If a l i n e i s perpendicular t o each of two I n t e r secting l i n e s a t their point o f intersection, t h e n it i s perpend i c u l a r to t h e plane of these l i n e s . Restatement; Let L, and Lp be l i n e s in plane E intersecting at A and let L be a l i n e through A perpendicular to L and L2. Then any line Lg In E through A is p e r p e n d i c u l a r t o L.
Proof: Reason
Statement 1.
2.
Let P be a point on L, 3, a p o i n t on L-,, Bg a point on Lp, and Bo a point on Ln, none of these points coinciding with A. Let Q be the o p p o s i t e to AQ
-;
AP.
1 1.
i n f i n i t e number of points.
on the ray 2. 1^oint such that
3.
In t h e plane containing L and 3 . Li, L, is the perpendicular b i s e c t o r of T^.
4.
B is equidistant from P and 6. [sec.
By the Ruler Postulate, each of these lines has an
4.
8-21
Point Plotting Theorem.
D e f i n i t i o n of perpendicular bisector (section 6 - 3 ) . Theorem 6 - 2 .
5.
Similar t o 3 and 4.
B3 is equidistant from P and Q.
7.
Steps 4 , 5 and 6, and Theorem 8-2.
In t h e plane containing L and L3, L3 is the perpendicular b i s e c t o r of
8.
Theorem 6-2.
9.
Definition of perpendicular
Bn is equidistant from P and
a. A is e q u i d i s t a n t from
P and Q ,
m.
bisector.
.O.
Definition of perpendlcularity of line and plane, since line in E through L3 I s
A.
Problem -S e t 8-2b
Suppose A, B and C are each equidistant from P and Q. Explain in terms of a definition or theorem why each p o i n t X of plane ABC is equidistant from P and Q.
, l q P
ex
A
I
41
B
Q
Ã
Explain the relationship between the line of Intersection L of two walls of your classroom and the plane of the floor. How many lines perpendicular to L could be drawn on t h e f l o o r ? Is L perpendicular t o every line that c o u l d be drawn on the floor?
Figure FRHB I s a square. _[ E. A I s not In plane FRHB. a. How many planes are determined by p a i r s of segments in t h e figure? Name them. b. At least one of the segments in this figure is perpendicular* t o one of the planes asked f o r In Part (a). Which segment? Which plane? A systematic approach t o such a problem is t o write down every pair of perpendicular segments you see in t h e figure. Then you can observe whether you have one line perpendicular to two intersecting lines.
4.
m.
A ABF is i s o s c e l e s w i t h B as vertex. AH = FH. 1 R is not in plane AFB. a. How many d i f f e r e n t planes are determined by the segments in the figure? Name them. b. Do you find a segment t h a t is perpendicular t o a plane? If so, tell what segment and what plane and prove your st at ernent
.
5.
- 1 plane
In t h i s figure, FB plane P, BR = EA.
*6.
P, and in A RAB, which lies in Prove A ABF ^ A RBF and L FAR Z L FRA.
Given the cube shown, with BR = BL. your answer is correct.
Does KR = KL?
Prove that
(Since we have not yet given a precise definition of a cube we state here, for use in your proof, the essential p r o p e r t i e s of the edges of a cube: The edges of a cube consist o f twelve congruent segments, related as shown In t h e p i c t u r e , such that any two intersecting segments are perpendicular. )
7.
<Ñ
In the accompanying figure WX is a line in plane E. Plane F % at Q. In plane F, % 1 ti?. IS la the intersection of E *-> a n d F. Prove RQ 1 E.
1
For all we know up to now t h e conditions s p e c i f i e d in t h e definition of a line and a plane perpendicular might be impossible to achieve. To reassure us, we need an existence theorem. The next theorem enables us to see that we are not talking about things that cannot exist In speaking of perpendicularity between lines and planes.
Theorem 8-4. Through a given p o i n t on a given l i n e there passes a plane perpendicular to the line.
Proof: Let P be a point on a line L. We show In six steps that there is a plane E through P perpendicular to L. 1. Let R be a p o i n t n o t on L. That there is such a point follows from Postulate 5a. 2. Let M be the plane determined by L and R. Theorem 3-3 3. 4.
5.
6.
tells us Let Q be there is L e t N be In plane
there is such a plane.
.
a p o i n t not on M P o s t u l a t e 5b assures us t h a t subh a p o i n t . the plane determined by L and Q. M t h e r e is a line L, perpendicular to L at P h he or em 6 - l ) , and in plane N there is a line Lp perpendicular to L at P. By Theorem 8-3, the plane I3 determined by L1 and Lp I s perpendicular to L at P.
1
If E L at P then e v e r y line In E and through P I s perpendie ular to L, by definition. May there be some l i n e s not in E but still perpendicular to L at P? The next theorem says, "?Jon. Theorem 8-5. If a l i n e and a plane are perpendicular, then the plane contains every line perpendicular to the given line at its point of intersection with the given plane. Restatement: If line L is perpendicular to plane E a t p o i n t P, and if M is a line perpendicular t o L at P, then M lies ¥1 E.
-
t Proof:
Reasons
Statements 1.
2.
L and M determine a plane F. Planes F and E intersect in a l i n e N.
3.
NIL.
3.
4.
M l L *
4.
= N. his means M and N are the same line.)
5.
M lies in E.
6,
1.
5* M
6.
2.
Theorem 3-4. Postulate 8. D e f i n i t i o n o f perpendicularity of l i n e and p l a n e . Given. M and N both l i e in plane I? by S t e p s 1 and 2, are both L by S t e p s 3 and 4, but Theorem 6-1 says there is only one such perpendicular. M = N by Step 5 and N lies in E by S t e p 2.
1
This theorem enables us to prove the uniqueness theorem that goes w i t h Theorem 8-4.
Theorem 8-6. Through a given point on a given line there is at most one plane perpendicular to t h e line.. Proof: 3 i n c e a perpendicular plane contains perpendicular l i n e s t h r o u g h the p o i n t , and since two d i f f e r e n t planes have only one l i n e In common (Theorem 3-41, there cannot be two such planes. Just as in a plane where t h e characterization Theorem 6-2 followedthe existence and uniqueness Theorem 6-1, so now we can prove a similar characterization theorem f o r space. Theorem 8-7. The perpendicular bisecting plane of a segment is t h e s e t o f all p o i n t s e q u i d i s t a n t from the end-points of t h e
segment. Note that this theorem, l i k e Theorem 6-2, has two parts. Restatement: Let E be the perpendicular bisecting plane of AB. Let C be t h e mid-point of Then (1) If P is in E, then PA = PB, and (2) If PA = PB, then P is in E. The proof is l e f t t o the student.
-
m.
Problem -S e t 8-2c a.
A t a p o i n t on a line how many lines are perpendicular t o
b.
t h e line? A t a point on a line how many planes are perpendicular to the line?
Planes E and F i n t e r s e c t in KQ, as shown in this f i g u r e . ¥
<->Â
w 1f at P and QP <-È why does PR lie in E?
G]E
^,
1* PR,
Assuming here t h a t AX = EX,
AY
=
BY,
EW, A 2 = BZ, AW =
why are W, X, Y and Z coplanar?
P l a n e E is the perpendicular bisecting plane of E,as shown I n the figure. a.
m
2 E2 a2 mLAFW =
AAKF
b.
2
Does FW = FK = FR?
Explain.
v 0
* 6 . Prove t h i s theorem: If L is a line intersecting plane E at p o i n t M, there i s at l e a s t one l i n e L f in E such that L 1 1 L. The next theorem is a lemma which is useful In proving l a t e r theorems. Theorem 8-8. Two lines perpendicular to the same plane are coplanar Proof: L e t l i n e s Li and L2 be perpendicular to plane E at t h e points A and B respectively. Let M be the mid-point of l e t L be the line In E which is the perpendicular b i s e c t o r of and l e t P and Q be two p o i n t s on L such that PM = QM* L e t C be a point on L, d i s t i n c t from A .
.
m, m,
P o s t u l a t e , A AMP 2 A AMQ, and so AP = AQ. Since L _[ E, L CAP and L CAQ are right angles, and the S.A.S. P o s t u l a t e g i v e s A CAP 2 A CAQ, so t h a t CP = CQ. From AP = AQ and CP = CQ it follows, by Theorem 8-7, t h a t C and A both l i e in the bisecting plane El of R. Hence, L, lies In E f In exactly the same way we prove t h a t Lp lies in E l . Hence, L, and L2 are coplanar. By the S.A.S.
.
Existence and Uniqueness Theorems. The following theorems cover a l l possible relations between a p o i n t , a l i n e and a perpendicular plane. They are s t a t e d here f o r completeness and for convenience i n reference. Theorem 8-9. Through a given p o i n t there passes one and only one plane pemendicular t o a given l i n e . Theorem 8-10. Through a given p o i n t t h e r e passes one and only one 1i n e perpendicular t o a given plane . The proof of each of t h e s e theorems has t w o cases, depending on whether or not t h e given p o i n t lies on the given line or plane, and each case has t w o parts, one f o r proving existence and one f o r proving uniqueness. T h i s makes a t o t a l of eight proofs required. Theorems 8-4 and 8-6 are two of these eight; the remaining six, some of which a r e hard and some easy, are given In Appendix VI. Theorem 8-10 assures us of the existence of a unique perpendicular t o a given plane from an external point. Hence, we are j u s t i f i e d In g i v i n g the following definition, analogous t o the one following Theorem 7-6, Definition. The distance t o a plane f r o m an external point is the length of the p e r p e n d i c u l a r segment from t h e p o i n t t o the plane. Theorem 8-11. The s h o r t e s t segment t o a plane from an external point is t h e perpendicular segment. The proof is similar to t h a t of Theorem 7-6. 8-3.
Review Problems
1.
Use a drawing if necessary to help you decide whether each statement is true or false. a. The Intersection of two planes may be a segment. b. If a l i n e i n t e r s e c t s a plane in only one p o i n t , there are at least two lines in the plane perpendicular to the line. c . For any f o u r points, there I s a plane c o n t a i n i n g them all. d. If three lines intersect in pairs, but no point belongs to all three, the lines are coplanar. e . It is p o s s i b l e f o r three lines to Intersect in a p o i n t , so that each is perpendicular to t h e other t w o , f . Only one l i n e can be drawn perpendicular to a given l i n e at a given point. g. A t a point in a plane there is only one line perpendicular
to the plane. h.
The greatest number of regions into which three planes can separate space is eight.
-1
2.
F r o m a p o i n t R outside plane E, RB E and I^B Intersects t h e plane in B. 'SK is any o t h e r segment from R, intersecting E in A . Compare the lengths arid RE. Compare the measures of L A and L B.
3.
If the goal posts at one end of a f o o t b a l l f i e l d are perpendicular to the ground, then they are coplanar even without a brace between them. Which theorem s u p p o r t s this conclusion? Can the goal posts still be coplanar even if they are not p e r p e n d i c u l a r to t h e ground? Could they fail to be coplanar even with a brace between them?
4,
Do there always exist a. two lines perpendicular to a given line at a given point on the line? b. two planes perpendicular to a given line at a given point on t h e line? c. two lines perpendicular to a given plane at a given p o i n t on the plane? d . two planes perpendicular to a given line? e . two intersecting lines each perpendicular to a given plane?
5.
The assumption t h a t two l i n e s L, and Lp are perpendicular t o plane E and L, and L2 i n t e r sect in point P not in plane E can be shown t o be f a l s e by
proving that the assumption leads to a c o n t r a d i c t i o n of a theorem about figures In a plane. Which theorem?
6.
<-Ã
Given MQ plane E, and 31 to plane E. How many d i f f e r e n t planes are determined by w[, <-È MW, WF and ?I^ Explain.
1
mlm.
AABF is i s o s c e l e s with vertex at 3. HF = HA. R is not In the plane AFB. a. How many d i f f e r e n t planes are determined by the segments b.
in the f i g u r e ? Explain. Locate and describe a line t h a t is p e r p e n d i c u l a r t o a plane.
8.
Given: P is in plane E which c o n t a i n s A, B, C; P I s e q u l distant f r o m A, B, C; line L 1 E at P. Prove: Every point, X, in L is equidistant from A, B, C
.
9.
10.
Given: Line L J^ plane A X a t Q; point P of L is equidistant from A , B, C . Prove: Every point of L I s e q u i d i s t a n t f r o m A, B, C. ( ~ i n t : Consider any point x # Q on L and show XA = XB = XC.)
Given:
AP l P ^ and D I E ;
11? a t
A
Q.
1
Prove; Ad S?. <-^ (~int: Take R on QC so that QB = O R , D r a w m ,
m.)
c
11.
Prove the following theorem: If from a p o i n t A outside a plane, a perpendicular and oblique (non-perpendicular) segments and AH are drawn, meeting the plane at unequal distances from 3, the segment which meets the plane a t t h e greater distance from 3 has the greater length.
A 3 1 plane
E. F and H are points of E such that BF > BH. Prove: AF > AH. Given:
12.
- + +
Prove t h a t each of f o u r rays AB,
AC,
+
AD and AE cannot be
perpendicular to the other three. <-9Ã
13.
<-*Â
<-Ã
1
Given: XB and YB are two l i n e s in plane E; m is a plane XI3 <-à <-È at B; n is a plane Y 3 at B; AB is the intersection of m and
1
n. Prove:
<-Ã
AB
j E.
Chapter 9 PARALLEL LIMES IN A PLANE
9-1. Conditions Which Guarantee Parallelism. Thus far in our geometry we have been mainly concerned w i t h what happens when l i n e s and p l a n e s I n t e r s e c t in c e r t a i n ways. We a r e now going to see what happens when t h e y do interseet It will turn o u t t h a t many more I n t e r e s t i n g things c a n be proved. W e f i r s t consider the case of two l i n e s . Theorem 3-3 gives us some information right away, since it says t h a t if t w o lines intersect they l i e in a plane. Hence, if two lines are not coplanar they cannot i n t e r s e c t . Definition: Two l i n e s which a r e not coplanar are said t o be skew. You can e a s i l y f i n d examples-' of skew lines in your classroom. This still leaves open the question as to whether two coplanar lines must always i n t e r s e c t . In Theorem 9-2 we shall prove t h e existence of coplanar lines that do not i n t e r s e c t , b u t are parallel, like t h i s :
.
-
L e t us first make a precise d e f i n i t i o n . Definition: Two l i n e s are parallel if they are coplanar and
.
do not; i n t e r s e c t Note that f o r two l i n e s t o be parallel two conditions must be s a t i s f i e d : they m u s t n o t intersect; they must both lie i n t h e same plane.
Theorem 9-1. Two parallel lines lie in exactly one plane. Proof: If L., and Lp are parallel lines it follows from t h e above definition that there is a plane E containing L-, and L p . If P is any point of % It follows from Theorem 3-3 that there is only one plane c o n t a i n i n g Li and P. Hence, E la the only plane containing L, and Lp.
We will use the abbreviation Ll 11 Ln to mean that the lines Li and LA are parallel. As a matter of convenience we will say t h a t two segments are parallel If the lines that contain them are parallel. We will speak similarly of a line and a segment, or a line and ray, and so on. For example, suppose we have given t h a t L-, \\ Lp, in the figure below;
+ @
4
+
z,
Then we can a l s o write It?%, AB 11 Lg, Ll 1 1 CD, BA 11 and so on. Each of these statements I s equivalent to the s t a t e -
ment that L, 11 Ln. It does n o t seem easy to t e l l from the d e f i n i t i o n whether two lines which seem to be parallel really are parallel. Every line stretches out i n f i n i t e l y far in two directions, and to tell whether two lines do n o t i n t e r s e c t , we would have to look at all of each of the two l i n e s . There i s a simple condition, however, which is s u f f i c i e n t t o guarantee t h a t two lines are p a r a l l e l . It goes like this:
-
-
Theorem 9-2. Two lines In a plane are p a r a l l e l if they a r e both perpendicular to the s m e line.
Proof: Suppose that Li and L2 are two lines In plane each perpendicular to a line L, at points P and Q.
There are now two possibilities: (1) L, and L2 Intersect in a point (2)
R.
L-, and Lp do not Intersect.
In Case (1) we would have two lines, L, and Lp, each perpendicular to L and each passing through R. This i s impossible by Theorem 6-1 i f R l i e s on L, and by Theorem 6-3 if R I s n o t on L. Hence, Case (2) i s the only possible one, and so, by definition, L, 11 Lp Theorem 9-2 enables us to prove the following important existence theorem.
E,
on L
Theorem 9-3, Let L be a line, and let P be a point not L. Then there is at least one line through P, parallel to
P r o o f ; Let Li be a line through P, perpendicular to L. (By Theorem 6-1, t h e r e is such a line. ) Lst L, be a l i n e through P , perpendicular to Li in the plane of L and P. By Theorem 9-2, L2 11 L. I t might seem natural, at this p o i n t , to try t o prove t h a t the parallel given by Theorem 9-3 is unique; that is, we might try to show t h a t in a plane through a given p o i n t n o t on a given l i n e there is only one p a r a l l e l to the given line. Astonishing as it may seem, t h i s cannot be proved on the basis of the postulates that we have s t a t e d so far; it must be taken as a new p o s t u l a t e . We will discuss t h i s in more d e t a i l in S e c t i o n 9 - 3 , In the meantime, before we g e t to work on the basis of this new postulate we shall prove some additional theorems which, like Theorem 9-2, tell us when two lines are parallel. We f i r s t give some definitions.
- 7 -
--
-----
A transversal of two coplanar l i n e s is a line which intersects them in t w o d i f f e r e n t p o i n t s .
Definition:
We say the two lines are "cut" by the transversal.
Definition: L e t L be a transversal of L-, and L2, intersecting them in P and Q. L e t A be a point of Ll and B a p o i n t of L2 such that A and B are on opposite sides of L. Then PW and WA are alternate L n t e ~ i o rawlea formed by the transversal to the two lines.
Notice that In the definition of a transversal, the two lines that we s t a r t with may or may not be parallel. B u t if they i n t e r s e c t , then the transversal is not allowed to intersect them at their common point. The situation in the figure below is not allowed:
That is, in thi3 figure L is a transversal to the lines Ll and Lp. Notice a l s o that a common perpendicular to two lines in a plane, as In Theorem 9-2, is always a t r a n s v e r s a l .
s,
Theorem If two l h e s are cut by a trmsver8a1, and if one pair of alternate I n t e r i o r angles are congruent, then the other pair of alternate I n t e r i o r angles are also congruent.
That I s , if / a = / a 4 , then / b E / b l . And if / b ~ / b t , The proof is left to t h e student. then / a ^ / a! The following theorem I s a generalization of Theorem 9-2, t h a t is, I t Includes Theorem 9-2 as a special case:
.
s.
Theorem Xf two lines are c u t by a transversal, and if a pair of alternate i n t e r i o r angles are congruent, then t h e lines are parallel.
L be a transversal to L-, and Lp, i n t e r secting them in P and Q. Suppose t h a t a pair of alternate i n t e r i o r angles a r e congruent. There are now two possibilities: Proof:
Let
(1)
L-, and Lp intersect in a point
(2)
L,
II
R.
L,*
In Case (1) the figure looks l i k e this:
L e t S be a point of Ll on the opposite side of L from R . Then / SPQ is an exterior angle of A P Q R , and / PQR is one of t h e remote i n t e r i o r angles. By Theorem 7-1, t h i s means that
m
/
SPQ 3 m
/
PQR.
-
B u t we know by hypothesis t h a t one p a i r of alternate interior angles are congruent. By the preceding theorem, b o t h p a i r s of
alternate I n t e r i o r angles are congruent. Therefore, m
/ SPQ = m /
PQR.
Since Statement (1) leads to a contradiction of our hypothesis, Statement (1) is false. Therefore Statement ( 2 ) is t r u e .
-
Problem Set 1.
a.
Does the definition of parallel lines s t a t e that t h e lines must remain the same distance apart?
b.
If t w o given lines do not l i e in one plane, can the lines be p a r a l l e l ?
, or
2.
Two lines in a plane are parallel if or if
3.
If two lines in a plane are Intersected by a transversal, are the alternate i n t e r i o r angles always congruent?
4.
In space, if two lines are perpendicular t o a t h i r d line, are the two lines parallel?
5.
a.
b.
6.
If the 80' angles were correctly drawn, would L, be parallel to Lp according to Theorem 9-5? Explain.
if
* L,
so*
4
B04
4
9
>Ln
How many different measures of angles would occur in t h e drawing? What measures?
In the figure, If the angles were of the s i z e indicated, which lines would be parallel?
At-, M1<
89'
SO-
>Â
90, 91.
M2<
^v
^3
1-2
v
v
491-
+
7.
Given a line L and a point F n o t on L, show how prot r a c t o r and ruler can be used to draw a p a r a l l e l to L through P.
8.
Suppose the following two definitions are agreed upon:
-
A vertical line is one containing the center of t h e
earth.
-
A horizontal line l a one which I s perpendicular to some
vertical line. Could two horizontal l i n e s be p a r a l l e l ? Could two vertical l i n e s be parallel? Could two h o r i z o n t a l l i n e s be perpendicular? Could two vertical
l i n e s be perpendicular?
Would every vertical line a l s o be horizontal? Would every horizontal l i n e also be vertical?
Could a h o r i z o n t a l line be p a r a l l e l to a v e r t i c a l line? Would every l i n e be h o r i z o n t a l ? I t possible t o f i n d t w o lines in space which are n e i t h e r
parallel nor i n t e r s e c t i n g ? 10.
Given;
DAB and
prove:
m
/
D
rnL ADC
=
BCD = 90?
#'
' \
m /BCD. m
C ¥s
AD=CB.
C a n you a l s o prove =
= 90,
=III/CBA
/
^,
ADO
A
\
00
X'^
, / '
'*Ã
'.,
0
11.
Given the figure with AR = RC = PQ, AP = PB = RQ, BQ = QC = PR. Prove : m / ~ + r n / ~ + m / ~ = l 8 0 . (~int: Prove m / a = m / A, m/b=m/B, mLc=m/C.)
12.
Given: A 3 = AC, AP = AQ. <¥à ÇProve: PQ 11 BC ( ~ i n t : L e t t h eb i s e c t o r of / A In t e r s e c t PQ a t R and BC at D.)
13.
Given: The figure with
.
/A=/B, AD = BC, AT = TB,
SD = SC. Prove :
B
9-2. Correspondin& Angles. In the figure below, the angles marked a and at
are
called corresponding angles:
Similarly,
c,
c!
and
and
b
d,
dt
are corresponding angles; and t h e pairs are also corresponding angles.
b1
Definition: If two lines are cut by a transversal, if / x and / y are alternate I n t e r i o r angles, and If / y and / z are vertical angles, then / x and / z are corresponding angles.
You should prove the following theorem.
-
Theorem 9-6. 1f two lines are cut by a transversal, and if one pair of corresponding angles are congruent, then t h e other three pairs of corresponding angles have the same property. The proof is only a little longer than that of Theorem 9-4,
u.
If two lines are cut by a tmmversal, and if Theorem a pair of corresponding angles are congruent, then the lines are parallel. The proof is l e f t to the student.
It looks as though the converses of Theorem 9-5 and Theorem 9-7 ought to be t r u e . The converse of Theorem 9-5 would say t h a t if two p a r a l l e l ' l i n e s are c u t by a transversal, then t h e alternate i n t e r i o r angles are congruent. The converse of Theorem 9-7 would say that If t w o parallel lines are c u t by a transversal, then corresponding angles are congruent. These theorems, however, cannot be proved on the basis of the postulates t h a t we have stated so far*. To prove them, we shall need to use the Parallel Postulate, which w i l l be stated i n the next section. The Parallel Postulate i s essential to the proofs of many other theorems of our geometry as well. Some of these you are already familiar with from your work in o t h e r grades. For example, you have known f o r some t i m e that the sum of the measures of the angles of any triangle i s 180. Yet, without the Parallel Postulate It is impossible to prove this v e r y Important theorem. L e t us go on, then, to the P a r a l l e l P o s t u l a t e .
9-3.
The Parallel
Postulate
.
Postulate 16. he Parallel Postulate .) Through 1 I1 a given external point there is at most one line I1 1 p a r a l l e l to a given line. Notice that we dontt need to say, i n the postulate, that there is a t least one such parallel, because we already know t h i s by Theorem 9-3. [see. 9-31
It might seem natural to suppose that we already have enough p o s t u l a t e s to be able to prove anything t h a t l a "reasonable"; and since the Parallel P o s t u l a t e is reasonable, we m i g h t try to prove it instead of c a l l i n g it a postulate. A t any rate, some very clever people felt this way about the postulate, over a period of a good many centuries. None of them, however, was able to find a proof. Finally, In the last century, I t was discovered t h a t no such proof is possible. The p o i n t is that there are some mathematical system that are almost like t h e geometry that we are studying, but not quite. In these mathematical systems, nearly all of the postulates of ordinary geometry are satisfied, but the Parallel Postulate is not. These on-~uclidean ~eometries" may seem strange, and In f a c t they are. (FOP example, in these "geometries" ---there is no such thin& as a aquare. 1 Not only do they lead to interesting mathematical theories, but they also have important applications t o physics. Now that we have the Parallel Postulate we can go on to prove numerous important theorems we could not prove without it. We s t a r t by proving the converse of Theorem 9-5.
--
-
Theorem 9-8. I f two parallel lines are cut by a transversal, then alternate i n t e r i o r angles are congruent. Proof: transversal
We have given parallel lines Ln, intersecting them in P
L, and and Q.
5,
and a
-
Suppose that / a and / b are not congruent. L e t L be a line through P f o r which alternate interior angles are congruent.
-
(BY the Angle Construction Postulate, there Is such a line.) L # Li, because / b and / c are not congruent. Now l e t us see what we have. By hypothesis, L, I[ L2. And by Theorem 9-5, we know t h a t L 1 f Lp. Therefore there are two lines through P, parallel to Lp. This is impossible, because It contradicts the Parallel Postulate. Therefore / a a / b , whloh was to be proved. The proofs of the following theorems are short, and you should write them f o r yourself: Then
u.
Theorem If two parallel lines are c u t by a transvernal, each p a i r o f corresponding angles are congruent.
Theorem 9-10. If two parallel lines are c u t by a transversal, I n t e r i o r angles on the same side of the transversal are supplementary.
Restatement:
/b to / e .
Lp. Prove t h a t supplementary
Ll 1 1 Lg and T I s supplementary to
Given
i n t e r s e c t s Ll and / d and /_ a I s
In a plane, two lines parallel to the same l i n e are parallel to each other. Theorem 9-11.
Theoremy-12. In a plane, if a line is perpendlcula~to one two p a r a l l e l lines it I s perpendicular to the other.
-
Problem S e t
Given : mL~=rnLB=rnLC=90. Prove: m / D = 90.
Prove that a l i n e parallel to the base of an isosceles triangle and intersecting the other two sides of the triangle forms another I s o s c e l e s triangle.
Given: Prove:
In the figure, RT = HS, % [I PQ = PT.
'8.
Q
T
4.
Review indirect proof as illustrated by the proof of Theorem 9-8. Give an Indirect proof of each of the following statements, showing a contradiction of the Parallel P o s t u l a t e .
a.
In a plane, if a third line M intersects one of two parallel l i n e s Li a t P, It a l s o
M
intersects the other
b.
In a plane, if a l i n e R intersects only one of t w o other lines Ll and Lp, then the l i n e s L, and L2 intersect Given;
Prove:
5.
a.
-
La -
.
-
R
intersects
R
does not i n t e r s e c t
Li Intersects
L,
at
P. L2.
L2
Prove: Two angles In a plane which have their s i d e s respectively parallel and extending both in the same ( o r both in opposite) d i r e c t i o n s are congruent. Given:
BA 1 1 ?? , BE [ I 3 ,
La
b.
Prove:
Two angles in a plane which have t h e i r s i d e s
r e s p e c t i v e l y parallel but have only one p a i r extending in the same d i r e c t i o n are supplementary.
Given:
--+
BA
B? Prove:
1[ 5 ,
11s. LABC2 / X Y Z . + rn&CYz (b) ~LABC
In ( a )
In
=
3.80.
(~ote: Only c e r t a i n cases are illustrated and proved here. All other cases can also be proved easily. The term "direction" I s undefined but should be understood.) 6.
Make drawings of various pairs of angles ABC and DEF such and 3. S t a t e a theorem that you think t h a t B? may be true about the measures of such angles.
1
*7.
5E 1
If Theorem 9-8 I s assumed as a postulate, then the Parallel Postulate can be proved as a theorem. { m a t is. it must be shown that there cannot be a second parallel to a l i n e through a p o i n t not on it.)
Given: Li and Lg are two l i n e s containing P, and Ll 11 M. Prove: t o M.
Lp not parallel
M
*8.
Show t h a t if Theorem 9-12 (1f a t r a n s v e r s a l is perpendicular to one of two parallel lines, it I s perpendicular to the other.) I s assumed as a postulate, the P a r a l l e l Postulate can be proved as a theorem. Glven: Ll 1 1 M Ln contain P. Prove:
and
Ll and (L, # L,.)
LA n o t parallel t o
9-4. Triangles. Theorem 3-13, The sum of the measures of the angles of a triangle is 180.
Proof : Given A ABC , l e t L be t h e line through B, parallel to AC. L e t / x , / x*, / y, / y t and / be as in the figure.
-
<->
=,
Let D be a point of L on the same side of AB a s C. <-Ã Since AC 11 A is on t h e same side of BD as C. Therefore C is in the i n t e r i o r of / ABD (definition of i n t e r i o r of an angle), and s o , by t h e Angle Addition P o s t u l a t e , we have m / A B D = m l z +m/yl. By t h e Supplement Postulate, m / X I + m / ABD = 180.
Therefore m / x l
+
+m/
z
+ /
m/y1
= 180.
m / x t and m / y = m / y f , because these are alternate I n t e r i o r angles. By substi-bufcion we g e t B u t we know by Theorem 9-8 that
m
x =
m/x+mLz+m/y=l80, which was t o be proved. From t h i s we g e t a number of Important corollaries: Corollary 9-13-1. Given a correspondence between t w o triangles. If t w o pairs of corresponding angles are congruent, then the t h i r d p a i r of corresponding angles are a l s o congruent.
The corollary says t h a t if
/
A
A'
and
/
B
S
/
B',
then / C S / C 1 . As the figure suggests, the corollary applies to cases where the correspondence given is n o t a congruence, as well as to cases where A ABC 9t A A t B I C t .
C o r o l l a r y 3-13-2. The acute angles of a right triangle are
complementary. Corollary 9-l3-3. FOP any triangle, the measure of am exterior angle is the sum of the measures of the two remote Interior angles.
--
Problem S e t 9-4 1.
2.
If the measures of t w o angles of a triangle are as follows, what I s the measure of the third angle? a.
37 and 58.
d.
r and
b.
149and30.
e.
45
c.
n
f.
90
and
n.
To find the distance from a point A to a distant point P, a surveyor may measure a small distance AB and also measure / A and / B. From t h i s information he can compute the measure of / P and by appropriate formulas then compute AP. If
/
87.5 and m compute rn / P.
m
3.
A =
/
3 =
+a and
a.
and
At.
45
-
a.
A
pp 0
88.3,
Why is the Parallel Postulate essential t o the proof of Theorem 9-13?
4.
On a drawing like the one on the right fill in the values of a l l of the angles,
5.
Given:
/
/ A s / X
and
can you correctly conclude that: 3
a.
a / Y,
LC
b.
SLZ?
Ei-
XY?
A.
A
6.
Given: BD and % 11 Prove:
7.
bisects
/
EBC ,
AB = BC.
The b i s e c t o r of an exterior angle at the vertex of an Isosceles t r i a n g l e is parallel to the base. Prove t h i s .
8.
Given:
The figure.
Prove:
s
+r
=
t
+ u.
( ~ i n t : Draw DB.)
*9.
Given:
/
BAG
and
is a right angle
QB = QA.
Prove:
*lo.
In the figure,
QB = QC.
Am,
Given: In / C is a r i g h t angle, AS = AT and BR = BT.
Prove:
m
/
STR =
c
45.
( ~ i n t : Suppose m / A = a. Write formulas in turn for t h e measures of other angles i n t h e figure In terms of a.)
A
9-5. Quadrilaterals in Plane. A quadrilateral is a plane figure with f o u r sides, like one of the following:
The t w o figures on the bottom illustrate what we might call the most general case, In which no two sides are congruent, no two sides a r e parallel, and no two angles are congruent. We can state the definition of a quadrilateral more precisely, in the following way.
Definition: L e t A, B, C and D be four p o i n t s lying in the same plane, such t h a t no three of them are collinear, and such t h a t the segments AB, BC, CD and DA i n t e r s e c t only In t h e i r end-points. Then t h e union of these four segments is a quadrilateral.
- - -
For short, we will denote this figure by ABCD. Notice t h a t in each of the examples above, with the exception of the l a s t one, the quadrilateral p l u s its i n t e r i o r forms a convex s e t , in t h e sense which was defined in Chapter 3. This I s not true of the figure at the lower right, but this figure is s t i l l a quadrilatera l under our definition. Notice, however, t h a t under our d e f i n i t i o n of a quadrilateral, figures like the following one are ruled out. [set- 9-51
Here t h e figure is n o t a quadrilateral, because the segments and DA i n t e r s e c t in a point which la not an end-point of e i t h e r of them. Notice also, however, that a quadrilateral can be formed, using these same four points as vertices, like t h i s :
Here
ABDC
Is a quadrilateral.
D e f i n i t i o m : Opposite s i d e s of a quadrilateral are two side3 that do not i n t e r s e c t . Two or Its pngles a r e opposLte if they do not contain a common side. Two s i d e s are called consecutive if they have a common v e r t e x . Similarly, two angles are called consecutive if they contain a common side. A diagonal is a segment joining two non-consecutive vertices.
and CD are opposite sides, CD or and are conare the diagonals of ABCD. Which consecutive?
In a quadrilateral ABCD, as are BC and AD and
AD. -
aecutive sides. AC and Which angles are opposite?
AB
Definition: A trapezoid l a a quadrilateral in which two, and only two, opposite sides are parallel.
Definition: A ~parallelopyamis a quadrilateral in which both pairs of opposite sides are parallel.
You should not have much trouble in proving t h e b a s i c theorems on trapezoids and parallelograms: Theorem $l-14. Either diagonal separates a parallelogram i n t o two congruent triangles. That is, if ABCD is a parallelogram, then A ABC s A CDA
.
Theorem 3-15. In a parallelogram, any two opposite sfdes are congruent.
Corollaq 9-15-1. If Ll 11 L, and If P and Q are any two points on Li, then the distances of P and Q from Lp are equal. This p r o p e r t y of parallel lines is sometimes abbreviated by saying t h a t " p a r a l l e l l i n e s are everywhere equidistant"
.
Definition: The distance between two parallel lines is the distance from any p o i n t of one l i n e to the other line. Theorem 9-16. are congruent.
In a parallelogram, any two opposite angles
-
Theorem 9-17. In a parallelogram, any two consecutive angles are supplementary. Theorem 9-18. The diagonals of a parallelogram b i s e c t each other.
In Theorems 9-14 through 9-18 we are concerned with several p r o p e r t i e s of a parallelogram; t h a t is, if we h o w that a quadrilateral is a parallelogram we can conclude c e r t a i n facts about It. In t h e following three theorems we provide for t h e converse relationship; t h a t is, if we know certain f a c t s about a quadril a t e r a l we can conclude t h a t I t is a parallelogram.
-
Theorem 9-19. Given a quadrilateral in which both pairs of opposite sides are congruent. Then the quadrilateral is a parallelogram. Theorern 9 - 2 0 . If two sides of a quadrilateral are papallel and congruent, then the quadrilateral is a parallelogram.
Theorem 9-21. If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. The f o l l o w i n g theorem s t a t e s two useful f a c t s .
this theorem is given In full.
The proof of
Theorem 9-22. The segment between the mid-points of two s i d e s of a triangle is p a r a l l e l to the third s i d e and half as long as the t h i r d side,
Res taternent : Given ^AN. L e t D and E be the mid1 AC points of and E. Then BE 11 AC , and DE = -y
.
Proof: Using the Point Plotting Theorem, let F be the Ñà p o i n t of the ray o p p o s i t e to ED such t h a t EF = DE. We give the rest of the proof in the two-column form. The notation f o r angles i s that of the figure, Statements 1. EF = ED.
Reasons F w a s chosen so as to make this t r u e . E is the mid-point of Vertical angles are congruent. The S . A . S . Postulate. Corresponding p a r t s of congruent triangles Theorem 9-5. AD = DB, by hypothesis, and DB = PC, by statement 4. Theorem 9-20. Definition of a parallelo-
e.
.
10.
1 AC. DE = 7
gram* 1 DE = -y DF, by statement 1, and DF = AC, by Theorem 9-15.
9-6.
Rhomb*a, Rectangle and Square.
Definitions: A rhombus is a parallelogram a l l of whose sides are congruent.
A rectangle i s a parallelogram a l l of whose angles are flight angles.
Finally, a square is a rectangle all of whose s i d e s are cohgruent,
As before, we leave the proofs of the following theorems f o r the student.
-
Theorem 9-23. If a parallelogram has one right angle, then it has four right angles, and the parallelogram I s a rectangle.
-
Theorem 9-24. In a rhombus, the diagonals are perpendicular to one another.
-
If t h e diagonals of a quadrilateral b i s e c t each other and are perpendicular, then the quadrilateral is a Theorem 9 - 2 5 ,
rhombus.
--
Problem S e t 9-6 1.
2.
For which of the q u a d r i l a t e r a l s -- rectangle, square, rhombus, parallelogram -- can each of the following properties be proved? a.
Both pairs o f opposite angles are congruent.
b.
Both pairs of opposite sides are congruent.
c.
Each diagonal bisects two angles.
d.
The diagonals b i s e c t each other.
e.
The diagonals are perpendicular.
f.
Each p a i r o f consecutive angles I s supplementary.
g.
Each pair of consecutive sides Is congruent.
h.
The figure is a parallelogram.
i.
Each pair of consecutive angles Is congruent.
j.
The diagonals are congruent.
With the measures of t h e angles as given in parallelogram
ABFH,
3.
give the degree measure of each angle.
In t h i s figure ABHQ and are parallelograms. What is the relationship of / M to /H? of / R to /H? Prove your answer. AFRM
A
B
F
4.
Would the following information about a quadrilateral be sufficient to prove it a parallelogram? a rectangle? a rhombus? a square? Consider each item of information separately.
Both pairs of Its opposite sides are parallel. Both pairs of I t s opposite s i d e s are congruent. 3
Three of its angles are right angles.
A
Its diagonals bisect each o t h e r .
Its diagonals are congruent. Its diagonals are perpendicular and congruent. Its diagonals are perpendicular b i s e c t o r s of each other.
It is e q u i l a t e r a l . It I s equiangular. It is equilateral and equiangular. Both pairs of Its opposite angles are congruent. Each p a i r of i t s consecutive angles I s supplementary.
5.
Given: ABCD is a parallelogram with diagonal AC. AP = RC. Prove: ogram.
6.
DPBR
I s a parallelA
Given: Parallelograms AFED and PBCE, a8 shown in t h i s plane figure.
Prove : ABCD ogram.
r.
0
is a parallel-
B
7.
If lines are drawn parallel to t h e legs of an Isosceles t r i a n g l e through a p o i n t in t h e base of the triangle, then a parallelogram is formed and Its perimeter is equal to the sum, of the lengths of the legs.
/\
Given: In t h e figure RS a RT, 11 E,
- PY 11 E. Prove:
a.
PXRY is a Parallelogram.
b.
PX
+ XR +
= RS
8.
+
RY
+ YP
I n this figure, if ABCD l a a parallelogram w i t h diagonals and %6 intersecting in Q and is drawn through Q, prove t h a t is bisected by Q.
Given the I s o s c e l e s trapezoid ABCD In which AD = GB and CD 11 E.
-
Prove
/
A S
/
B.
T
P
RT.
AC
9.
S
Q A
F
8
10. The median of a trapezoid is the segment Joining the midpoints of its non-parallel sides. a.
Prove the following theorem: The median of a trapezoid is parallel t o the bases and equal in l e n g t h to half the sum of the l e n g t h s of the bases.
Given:
Trapezoid
ABCD
w i t h CD 11 E , P the midpoint of AD and Q the midpoint of BC.
PQ 11 AB ( A 3 + cD). PQ = 2 Prove:
-
( ~ i n t : Draw
AB
b.
c.
11.
at
<-Ã
DQ meeting
K.)
If AB = 91n. and DC = Tin, then PQ =
A
B
1 If DC = % and A 3 = 7 , then PQ =
convex quadrilateral with vertices labeled consecutively ABCD is called a k i t e if AB = BC and CD = DA. Sketch some k i t e s . S t a t e as many theorems about a k i t e as you can and prove at l e a s t one of them. A
Q u a d r i l a t e r a l ABCD with P, Q, R, S the midpoints o f the s i d e s . Given:
Prove:
RSPQ is a parallelogram, and PR and SQ b i s e c t each other.
13+ Given: Int h e figure AD < BC, DA 1E ,
Prove:
m
/
C
D.
*l4. Prove that
t h e sum of the lengths of the perpendiculars drawn from any p o i n t in the
base of an Isosceles triangle to the legs is equal t o the length of the altitude upon e i t h e r of the legs.
(~int: Draw
% 1 E.
Then the figure suggests t h a t PX and Q3' are congruent, and that and % are congruent. )
-
*l5. Prove t h a t the sum of the lengths of
t h e perpendiculars drawn
from any point in the i n t e r i o r of an equilateral triangle to the three sides is equal t o the l e n g t h of an altitude.
(Hint: Draw a segment; perpendicular to the altitude used, f r o m the interior p o i n t .) 16. Given a hexagon as In the figure w i t h AB 11 OC,
BC DE
[OD, CD 11 E , 11 OF, EF 1 c. -
Prove:
FA
I[
m.
A
F
E
D
C
17. a .
Given
sl, BBt , CC1
-
-
are p a r a l l e l and 11 A f B ' , BC I ~ B ' C I as in figure. prove: b.
AC J J A ' G ~ .
A B
Is the figure necessarily a plane f i g u r e . Will your proof apply If it
18. Given A3CD is a square and the p o i n t s K, L, M H d i v i d e the s i d e s as shown, a and b being lengths o f the indicated segments.
Prove : KLMN
*19. Show that if i n t e r i o r of
is a square.
ABCD is a parallelogram then D / A3C.
is in t h e
*20. Show that the diagonals of a parallelogram i n t e r s e c t each other
.
9-7.
-
-
Transversals To Many Parallel Llnes.
D e f l n i t i o n s : If a transversal i n t e r s e c t s two lines Ll, L2 in points A and 3, then we say that Ll and L2 intercept the segment AB on t h e transversal.
Suppose that we have given three Hnes Ll, L2, L3 and a t r a n s versal intersecttng them in p o l n t s A , B and C. If AB = BC, then we say that the three line3 3mtercept congruent segments on the transversal.
We s h a l l prove the following:
-
Theorem 9-26. If three parallel l l n e s intercept congruent segments on one tramversal, then they Intercept congruent segments on any o t h e r t r a n s v e r n a l .
Ll, L2 and L3 be p a r a l l e l lines, cut by a t r a n s v e r s a l TI in points A , B and C. Let T2 be another t ~ a n s v e r s a l , c u t t l n g these lines in D, E , and F. We have Proof:
Let
given t h a t AB = BC;
and w e need t o prove that
DE = EF. We w i l l first prove the theorem for the c a s e in which and T2 are n o t p a r a l l e l , and A # D, as In-the figure:
-
Let
T3 be the line through
Tl
p a r a l l e l t o T2, htwsecting L2 and L3 in G and H; and l e t T4 be the l i n e through 3, p a r a l l e l to T2, intersecting L3 in I. L e t x, y, w and z be as i n d k a t e d in the figure.
1
A,
Statements
Reasons
L x S L z .
1. 2.
= BC.
911 T ~ LwzLy.
4. 5.
BCI.
ABQ AG =
3.
Theorem 9-9. Hypothes I s , Theorem 9-11. Theorem 9-9. A. S . A.
6. D e f i n i t i o n of congruent
BI.
triangles. AGED and BIFE papal lelograms
.
AG =
DE
a d
are
BI
=
El?.
DE = EF.
7.
Definition of parallelogram3.
8.
Upposite s i d e s of a parallel-
ogram are congruent. 9. Steps 6 and 8.
This proves the theorem for the case Ln which the two trans-
versals are not parallel, and intersect Ll In two different points. (1)
The other cases are easy.
If the two tmnsversals are p a r a l l e l , llke T2 and T3
in the f i g w e , then the theorem holds, because opposite siaes of a p a r a l l e l o g r m are congruent. ( ~ h u s ,if AG = I%, It follows t h a t
DE
=
EF.)
If the two transversals Lntersect at A , 1lke !TI and T3 Ln the figure, then the theorem holds; In fact, we have already proved that If AB = BC, then AG = GH. (2)
he f o l ~ ~ w i ncgo r o l l a r y generalizes
meorem 9-26.
Corollary 9-26-1. If three or more parallel line8 i n t e r c e p t congruent segments on one t r a n s v e r ~ a l ,then they i n t e r c e p t congruent segments on any other transver~al.
That I s , given t h a t
BIB2 = B2B3
=
B3B4
=
. . .,
on, This follows by repeated applications of the theorem t h a t we have j u s t proved.
and so
DefInTtion: Two or more sets are concurrent if there l a a p o i n t which belongs to a l l of the sets.
In particular, three or mom llms are concurrent if they a l l pass through one p o i n t .
The following theorem is an interesting application of
Co~ollary9-26-1.
-
Theorem 9-27, The medians of' a trlangle are concurrent in a point two-thirds the way from any vertex to the mid-point of the opposite side. Given: In AABc, D, E and F are the mld-point3 of and AB respectively.
c
iz,zi
To Prove:
There I s a p o i n t
-
-
which lies on AD, BE CF; and AF = VJ2 AD,
P
and A
F
B
Sketch of proof:
(1)
Ll, L2, L3, parallel lines d i v t d b g Let
(b)
e
LQ and L with L3 = AD be five 5' i n t o f o u r congruent segments, Then
Ll, L2, L3, Lh divide BE into three congruent sewents, and so if P 1s t h e p o i n t of i n t e r s e c t i o n of 2 AD and E, then BP = 3 BE.
If
In t h e same way, w i t h l i n e s parallel to PI 1s the i n t e r s e c t i o n o f BE and &,
OF,
we f i n d t h a t then BPI = $ B E .
Pv
P,
(3)
From (1) and (2) and Theorem 2-4 it follows t h a t and t h e r e f o r e the three medians are concurrent.
(4)
Since we now know that passes through P we can easily 2 g e t CP = -7 C F from the figure in (1), and similarly g e t AP = 7 AD from the figure in ( 2 ) .
=
'
The centroid of a triangle I s the point of concurrency of the medians.
Definition:
Problem Set Q-7
a.
Prove <-È
b.
Do AC,
ZY
= YX. <->
<Ñ
TR and ZX have to be coplanar* to carry out the proof?
The procedure at the right can be used to rule a sheet of paper, 3, I n t o columns of equal width. If A is an ordinary s h e e t of ruled paper and B is a second sheet placed over I t as shown, explain why
3.
Divide a given segment following method:
(1) D r a w ray (2)
+ AR
-
-
-
The medians of
at
AB. )
-
N3N4
and
N N
4 5
AN,,
of any convenient length.
Measure AN? and u s e your p r o t r a c t o r t o draw corresponding angles congruent to / AN B with vertices a t 5 Explain why
4.
( n o t c o l l i n e a r with
Use your ruler t o mark o f f congruent segments
N1N2, N g M 3 3 (3) Draw NcB. (4)
i n t o f i v e congruent parts by t h e
AB
is divided into congruent parts.
AABC meet
Q, as shown in this
figure.
If BF = 18, A Q = 10, CM = 9, then BQ = GH = , CQ =
it-
J
In equilateral
AABCif
one median is 15 Inches long, what I s the distance f r o m the centroid to A? To the midp o i n t of TO side
AB?
AC?
-
Given: CM b i s e c t s M. % bisects ?% a t
at P.
Prove: Q is a trisection point of E; t h a t is, AQ = 2QC.
(Hint: On the ray opposite 4 to CB take p o i n t E such
c
t h a t CE = CB and show t h a t % is contained in a median of &BE, )
What is the smallest number of congruent segments i n t o which AC can be divided by 8ome set of equally spaced parallels which will include the parallels ?% and e 8 CT if:
4
A
bR
\
IS,
c.
and BC = I? 1 A 3 = +arid BC = I ? AB = 21 and BC = 67
e.
A 3 = y^" and
a. b.
AB=2
BC = I?
4
B
<
\c
@S *T
*8.
Prove t h a t t h e l i n e s through opposite vertices of a parallelogram and the m i d p o i n t s of t h e opposite s i d e s trisect a diagonal.
( ~ i n t : Through an extremity of the diagonal, consider a parallel to one of the lines .) GLven:
ABCD I s a parallelogram. X and Y are midpoints.
Prove:
AT = TQ = QC.
Review Problems 1.
I n d i c a t e whether each of the following statements I s true in ALL c a s e s , true in SOME cases and f a l s e in others, o r true in NO case, using the l e t t e r A , S o r N:
a
Line segments In t h e same plane which have no p o i n t in common are parallel.
b.
If two s i d e s of a quadrilateral ABCD then A3CD is a trapezoid.
c.
Two angles in a plane which have t h e i r s i d e s respectively perpendicular a r e congruent.
d.
If two parallel lines are c u t toy a transversal, then a p a i r of alternate exterior angles are congruent.
are parallel,
If two lines are c u t by a transversal, then the rays bisecting a pair of alternate interior angles are parallel.
In a plane, if a line is parallel to one of l i n e s , it is parallel t o the other.
two parallel
In a plane two lines are e i t h e r p a r a l l e l or they intersect
.
I n a parallelogram the opposite angles are supplementary. The diagonals of a rhombus b i s e c t each other. All three exterior angles of a triangle are acute. A quadrilateral having two opposite angles which are
right angles I s a rectangle.
The diagonals of a rhombus are congruent.
If a quadrilateral is equilateral, then a l l of its angles are congruent. If two opposite sides of a quadrilateral are congruent and the o t h e r two sides are parallel, the quadrilateral Is a
parallelogram.
The diagonals of a rhombus bisect the angles of the rhombus.
If the diagonals of a parallelogram are perpendicular, the parallelogram i s a square. If a median to one side of a triangle is not an altitude, the other t w o sides are unequal in length. E i t h e r diagonal of a parallelogram makes fcwo congruent triangles with the s i d e s .
If a diagonal of a quadrilateral divides it Into two congruent triangles, the quadrilateral Is a parallelogram If t w o lines are intersected by a transversal, the alternate interior angles are congruent.
u
.
v.
w
.
x. 2.
A l l four sides of a rec tangle are congruent. A l l four angles of a rhombus are congruent. A
square is a rhombus
.
A square is a rectangle.
Would the following information about a quadrilateral be sufficient t o prove it a parallelogram? A square? A rhombus? A rectangle? Consider each item of Information
separately.
I t s diagonals bisect each o t h e r
Its diagonals a r e congruent
.
.
It is equilateral. It is e q u i l a t e r a l and equiangular. A diagonal b i s e c t s two angles.
Every two o p p o s i t e sides are congruent.
Some two consecutive sides are congruent and perpendicular. The diagonals a r e perpendicular. Every two opposite angles are congruent.
Each diagonal b i s e c t s two angles. Every
two consecutive angles are supplementary.
Every two consecutive sides are congruent.
and
/
B have t h e i r sides respectively parallel.
If only one pair of corresponding sides extend in the same direction the angles are If corresponding sides extend in opposite directions, then the angles are
.
In Problems 4, 5 and 6 below select the one word or phrase that makes the statement t r u e . 4.
The bisectors of the opposite angles of a non-equilateral
(a) coincide, (b) are perpendicular, intersect but are not perpendicular, (d) are parallel.
parallelogram (c)
5.
*6.
7.
The figure formed by Joining the consecutive mid-points of the sides of a rhombus I s (a) a rhombus, (b) a rectangle, ( c ) a square, ( d ) none of these answers.
The figure formed by joining t h e consecutive mid-points of the sides of quadrilateral ABCD is a square (a) if, and o n l y if, the diagonals of ABCD are congruent and perpendicular, (b) if, and only if, the diagonals o f ABCD are congruent, ( c ) I f , and only If, ABCD is a square, (d) if, and only if, the diagonals of ABCD are perpendicular. In the left-hand column below, certain conditions are specif l e d . In t h e right-hand column, some deducible conclusions are l e f t for you t o complete.
diagona 1s of
MKWR
.
Conditions :
Conclusions:
a.
MKWR
is a parallelogram, m/a = 30. and
m/d =
and ~ / R W K =
b.
MKWR
is a rectangle and m/a = 30. MKWR Is a rhombus, m/a = 30 and MK = 6 .
m/d =
and m/b =
m/b =
and BK
c.
=
A
8.
In the figure AE=EB, GF=8, <-*Â <-Ã CF = FB, DE 11 CB. Given:
Find:
9.
10.
DO.
F
C
8
If the perimeter (sum of lengths of s i d e s ) of a triangle I s 18 inches, what is the perimeter of the t r i a n g l e formed by Joining the mid-points of sides of the first triangle? a.
If m / A = 3 0 and m/ C = 25, w h a t is t h e measure of / CBD?
b.
If r n / A = a and 0 / . c s 7a, what is
m/ CBD?
C
A
m/ ABC?
>
D
A
11.
Show that the measure of / E, formed by the bisector of / ABC and the b i s e c t o r of exterior / ACD of A ABC, I s 1 equal to -y 4 A .
E
B
D
< - ^ < - > +
>B
9
12. In the figure AB 11 CD, EG b i s e c t s / BEP, m/ G = 90. If the measure of / GEF = 25, what is the measure of / GPD?
6
Y
4
-
13.
-
D
\F
Given: AB and CD which bisect each o t h e r at 0.
- -
Prove:
AC
1
BD. C
14.
B
Given: ABCD i s a parallelogram w i t h diagonals AC and 1 AC DB. AP = RC < -y
-
Prove:
DPBR
is a
parallelogram.
15.
Prove or disprove:
If a quadrilateral has one pair of parallel sides and one pair of congruent sides, then the quadrilateral is a parallelogram.
-
AM I s congruent t o AABCI s a r i g h t t r i a n g l e .
*16. In AABC,
17. Prove:
median
-
MC.
Prove t h a t
If t h e b i s e c t o r s of two consecutive angles of a parallelogram i n t e r s e c t , they are perpendicular to each o t h e r .
A
Given: ABODE I s a <È *à as shown. AE 11 CD. P is mid-point of K is mid-point of Prove:
19.
KE
bisects
E
pentagon AE = CD. AB BC
-. 7
.
PM .
When a beam of light is reflected from a smooth surface, the angle between the incoming beam and the surface is congruent to the angle between the reflected beam and the surface.
In the accompanying figure, m/ ABC = 90, m / BCD = 75, and the beam of light makes an angle o f 35' with E . Copy t h e figure and complete the path of t h e light beam as it r e f l e c t s from E , from E, from E, and from AB again. A t what angle does the beam r e f l e c t from AB the second time?
20.
.
Given t~iangle ABC wlth and medians If is extended its own length to D, and CS is extended its o m length to F, prove that 3?, B and ll as shorn are collinear.
Chapter 10
10-1.
Parallel Planes.
D e f i n i t i o n : Two planes, or a plane aria a line, are p a r a l l e l if they do not i n t e r s e c t , If planes El arid E2 are parallel we w r i t e El 11 E2; if line L and plane E are parallel we w r l t e L 11 E or I3 1 1 L. As we will soon see, parallels in space behave in somewhat the same way as parallel lines Ln a plane. To study them we do not need any new postulates. However, in s p i t e of the similarities I t is n e c e s s a q , 3.n studying theorem and t h e i r p r o o f s in t h i s chapter, to distinguish carefully between parallel llnes and parallel planes. Two parallel plane3 such as E and F in the f i r s t Figure below contain lines such as Ll and L2 which a m n o t parallel And the second Figure shows parallel l i n e s MI and M2 l y i n g in intersecting planes G and H.
-
The following theorem d e s c r l b e ~a common situation in whlch parallel planes m d parallel lines o c c u r in the same F i g u r e .
-
Theorem 10-1. If a plane Intersects two parallel planes, then tt intersects them in two parallel lines
.
Proof: Glven a plane E, i n t e m e c t i n g two parallel planes El and E ~ . By Postulate 8, the b t e r s e c t ~ o mare lines Ll and L2. These lines are In the same plane E; and they have no point In cornon because El and E2 have no poLnt in comon. Therefore, they are parallel by the definition of parallel l i n e s .
-
Theorem 10-2. If a line is perpendicular to one of two parallel planes, it is perpendicular to the other.
Proof: L e t planes El and E2 be parallel and let llne L be perpendicula~to El. In E2 t a k e a p i n t A n o t on L, and l e t l3 be the plane determined by L and A . By the preceding theorem E i n t e r s e c t s El and E2 in parallel lines Ll and L2. L Ll since L Bl, and so by Theorem 9-12 (look it up) L L2. Now take a point A1 In E2 not on L2 and repeat the process. We thus obtatn two llnes in E2 each perpendZcular to L, and so L E2, by !?heorem 8-3.
1
1
1 Theorem 10-3.
Two planes p e r p e n d i c u l a r to t h e same line aye
parallel.
Proof:
1
The figure on the l e f t shows w h a t happens when
-
El L at P and E2 1 L at Q: we wish t o show El 1 1 E2. If El and E2 are n o t parallel, they i n t e r s e c t . Let R be a 4+ e common point. Consider the lines PR and a. Then L 1 PR a d L % because L i8 perpendlcula~to every line in El through P and every line in E2 through Q. ThZs glves two perpendiculars to a line from an eiternal p o i n t , which is impossible, by
1
C o r o l l a r y 10-3-1, If two planes are each parallel to a t h i r d plane, they are parallel to each other.
Proof: L e t El 11 E3, E2 1 1 E3. Let L be a line perpendicular to E3. By Theorem 10-2 L El and L E2. Thus El and E2 ape each perpendicular t o L and El 11 E2 by the Theorem 10-3.
1
1
-
Theorem 10-4. Two l i n e s p e r p e n d i c u l a r t o the same plane are parallel.
proof: By meorem 8-8 two such lines are coplanar* Since they are perpendicular to the given plane, say at p o i n t s A and B, they are perpendicula~ to parallel.
s.Hence by Theorem 9-2 they
aye
C o r o l l a r y 10-4-1. A plane perpendicular to one of two parallel lines I s perpendicular to the other.
Proof: L e t Ll 1 1 L2, Ll E m L e t L3 be a 1Lne perpendicular to E through any p o l n t A of L2. L~ exists by Theorem 8 - 9 . Then by Theorem 10-4 Ll 11 L3. Hence, by the Parallel Postulate L3 = L2, and 80 L2 E.
1
Corollary 10-4-2. If two ltnes are each parallel to a t h i r d they are parallel to each other. Proof: Let Ll 11 L2, Ll [I La. Let E p e n d i c u l a r t o Ll. By the above corolla^ E and s o by t h e above theorem 11 L3.
1
be a plane p e r L2 and E L3,
1
Theorem 10-5. Two parallel planes are everywhere equidistant. That is, a l l segments perpendicular to the two planes and having t h e i r end points in the planes have the same length.
Proof;
Let
and
be perpendicular segments between
t h e p a r a l l e l planes E, and Ep. By Theorem 10-2, each of the segments Is perpendicular to each of the planes. By Theorem 10-4, ¥fr PQ 11 and this means, In particular, t h a t and % lie # *-> i n the same plane ET By Theorem 10-1, QR 11 PS. Therefore, PQRS is a parallelogram. Opposite sides of a parallelogram are congruent. Therefore, PQ = RS, which was t o be proved. (obviously PQRS is a rectangle, but this f a c t does not need to be mentioned in the proof .)
%;
?S
Problem Set 10-1 1.
D r a w a small s k e t c h to illustrate t h e hypothesis of each of the following statements. Below each s k e t c h indicate whether the statement is true or false.
If a line is perpendicular t o one of two parallel planes it is perpendicular to the o t h e r . Two l i n e s parallel to the same plane may be perpendicular to each o t h e r .
Two planes perpendicular to the same line may intersect. If a plane intersects two i n t e r s e c t i n g planes, the l i n e s of intersection may be parallel.
If t w o planes a r e both perpendicular to each of two parallel lines, t h e segments of the two lines i n t e r c e p t e d between the p l a n e s are congruent. If two planes, perpendicular to the same line, a r e i n t e r s e c t e d by a t h i r d plane, the lines of Intersection are parallel.
If a line lies in a plane, a perpendicular t o the line is perpendicular to the plane.
If a line lies in a plane, a perpendicular to the plane at some point of the line is perpendicular to the line.
If two lines are parallel, every plane containing only one of them Is p a r a l l e l t o the other line. .
.
If two lines are parallel, every line Intersecting one of them i n t e r s e c t s the other.
If two planes are parallel. any line in one of them is parallel to t h e o t h e r .
If two planes are parallel, any l i n e I n one of them i s p a r a l l e l to any line in the other.
2.
Given lines L, and Ly intersecting parallel planes m, n, and p at points A, B, C, and
Y,
X,
with
2,
B the mid-point of Prove : XY = YZ .
3.
Given: plane A 3 1 r. OX CY s
=
plane s . Prove : AX
4.
Given:
A,
11
plane in
= AY.
C
in
z.
m;
r,
Given:
In the figure m
Prove:
AD
=
11 n,
CB.
Planes E and F are <-> perpendicular to A , . <-> <-> L i n e s BK and BH, in plane F, determine <-à with AB two planes which I n t e r s e c t E in <-> ¥(AD and A C . Certain lengths a r e given, as in the figure.
and BACH description of them? length of CD? Are
BKDA
Can you give a further Is ABHK Â ^\CAD? C a n you give the
parallelograms?
In the figure half planes n and m have a common <-Ã edge AB and i n t e r s e c t < parallel planes s and t < Â ¥ Ã ˆ Â ¥ < - >
in l i n e s AD, AE, and as shown,
BG,
BP
Prove t h a t
/
DAE
S
/
GBF.
Show how to determine a plane containing one of two skew lines and parallel to the other. Prove your construction.
Given: PL and PM in plane E . RL 1 SM MP. RL 11 SM.
1
Prove: (flint:
RL At
lie
1 E, SM 1 E. P draw
10-2. Dihedral Angles, Perpendicular Planes.
We have considered perpendicularity between two lines, and between a line and a plane. We have y e t to define perpendiculari t y between two planes. This can be done in various ways, and we choose the one t h a t has the closest analogy with the definition of perpendicular lines.
Definitions: A dihedral angle is the union of a line and two non-coplanar half-planes having t h i s l i n e as t h e i r common edge. (Compare wlth the definitLon of angle in Chapter 4,) The line is called the e d ~ eo f the dihedral angle. The union of the edge and either half-plane I s called a -3face or s i d e , of the dihedral angle.
-
<->Â
and B points on d i f f e r e n t s i d e s , we denote the dihedral angle by /A-PQ-B.
If PQ is the edge, and A
Analogous to t h e discussion on page 88 we see t h a t two i n t e r s e c t i n g planes determine four dihedral angles.
Terms such as v e r t i c a l , interior, e x t e r i o r , e t c . can be applied to dihedral angles. Definitions of these terms can be considered an exercise f o r the student. To d e f i n e r i g h t dihedral angles, however, we need to t a l k .-aboutthe measure of a dihedral angle. One might a t f i r s t think that we must introduce four new p o s t u l a t e s , analogous to those In Section 4-3. However, t h i s is not necessary, for we can r e l a t e each dihedral angle with an o r d i n a r y angle, as follows:
Definition: Through any point o n t h e edge of the dihedral angle pass a plane perpendicular to t h e edge, intersecting each of the aides i n a r a y . The angle formed by these rays is called a plane angle of the d i h e d r a l angle. The sides of the plane angle are perpendicular t o the edge o f the dihedral angle, so another way of defining the plane angle would be t h e angle formed by two rays, one in each s i d e o f t h e dihedral angle, and perpendicular t o its edge at the same point.
It is natural a t t h i s p o i n t to use the measure of the plane angle as a measure o f the d i h e d r a l angle, but b e f o r e we do this we must prove t h a t any two plane angles of a dihedral angle have
the same measure.
-
Theorem 10-6. Any two plane angles of a given dihedral angle are congruent,
Figure B .
Figure A .
L e t V and S be the vertices of two plane angles of /A-PQ-B. (~igureA . ) On the s i d e s of / V take points U and W distinct f r o m V. On the s i d e s of / S take p o i n t s R and T such that S R = W , ST = W . (Figure B. ) VU and are coplanar and perpendicular to PQ; hence they are parallel by Theorem 9-2. Hence by Theorem 9-20 (look i t up) WRS I s a parallelogram and UR = VS and UR 11 Similarly, WT = VS and 11 m. Hence UR = WT and 11 the latter f a c t following from Corollary 10-4-2. URTW is thus a parallelogram, and UW = RT. It follows f r o m the S .S .S Theorem t h a t A U W at A RST, and so rn/ UW = m/ RST.
Proof:
-
-
E.
m,
.
Thus we can make the following definitions.
D e f i n i t i o n s : The measure o f a dihedral angle is the real number which is the measure o f any of I t s plane angles. A dihedral angle is a right dihedral angle If its plane angles are ~ i g h t angles. Two planes are perpendicular if they determine right dihedral angles.
The following are some immediate consequences of these d e f i n i t i o n s . Their proofs are left as exercises. Corollary 10-6-1. If a line is perpendicular to a plane, then any plane containing this line is perpendicular to the given plane.
Given: AB.
E,
<->
F
Prove:
F contains
1E.
(~int: Take E.)
<-Ã
BC
1 <-> PQ
in
Corollary 10-6-2. If two planes are perpendicular, then any line in one of them perpendicular to t h e i r line of intersection, is perpendicular to the o t h e r plane.
El
(~intt In the above figure; given w E. Take BC as before.) Problem -S e t 10-2
1
.
~ ; m e the 8 ix dihedral angles in t h i s three dimensional
figure
.
F
1 E, AS 1z;prove
2.
E a c h o f G , ?k and is perpendicular to the o t h e r t w o . a = b = c = 45. What I s the measure of / C-PA-B? of / CAB?
3.
Draw a small sketch to i l l u s t r a t e the hypothesis of each of the following statements. Then i n d i c a t e whether each is True (1) o r False (0).
If a plane and a line not in it are both perpendicular to the same l i n e , they are parallel to each other.
If a plane and a line not in it are both parallel to the same line they are parallel to each other.
If parallel planes E and F are cut by plane lines o f i n t e r s e c t i o n are perpendicular.
Q,
the
I f two planes are p a r a l l e l t o the same line they are parallel to each other.
Two lines parallel to the same plane are p a r a l l e l to each other.
of p a r a l l e l lines intercepted between two parallel planes are congruent. Segments
If planes E and F are perpendicular to they intersect I n line %$.
<-È
AB,
then
Two planes perpendicular to the same plane are p a r a l l e l to each o t h e r . Two lines perpendicular to t h e same line at the same point are perpendicular t o each o t h e r .
plane perpendicular t o one of two intersecting planes m u s t i n t e r s e c t the other. A
[sec. 10-21
k.
If two i n t e r s e c t i n g planes are each perpendicular to a t h i r d plane, their line of intersection is perpendicular t o t h e t h i r d plane.
Prove: If two intersecti n g planes are each perp e n d i c u l a r to a third plane, t h e i r i n t e r s e c t i o n is perpendicular to that
\
x/
third plane.
Y
Given: Planes r and s intersect in PQ (P being chosen f o r convenience on plane E) . r E and s E.
1
1
Prove:
QP 1 E.
-
fl 1 E ,
5.
<-È
(I-iint: In plane E, draw and use Corollary 10-6-2.)
<->
CD and FK are perpendlcul a r to plane E. Other given information is as shown in the figure.
Which t w o segments have the same length?
t
1 DC
and
'6.
Prove t h e following theorem: If three planes El, Ey and E, intersect in pairs and determine three lines Li2, Ln and Lgn, then e i t h e r t h e three lines are concurrent or each p a i r of the lines are parallel.
( ~ i n t ; The figure shows E, and Eg meeting in Lip. If En 11 LIP will the three lines L^*, L13 and Lgg, be concurrent or parallel? Give proof. If En i n t e r s e c t s Lip In some point P will the three lines be conc u r r e n t or parallel? Give proof. )
-7.
Desarppesr Theorem.
@ ,'
Lie
If two triangles lying in non-parallel
planes are such that the lines joining corresponding vertices are concurrent, then if corresponding side-lines i n t e r s e c t , their p o i n t s o f i n t e r s e c t i o n are collinear.
AA'B'
Given the triangles AABCand C' in non-parallel planes such t h a t , and CC! i n t e r s e c t <-> at U. L e t t h e lines ^S and meet at X, CA and <Ñ C I A 1 meet at Y, and AB and AtB1 meet a t 2. Prove t h a t the p o i n t s X, Y, 2 l i e on a line.
Restatement.
-
-
z*st i'B^
.
Projections You are familiar with a slide p r o j e c t o r which projects each point of a slide onto a screen. Each figure in the s l i d e is projected as an enlarged figure on the screen. In t h i s s e c t i o n you will n o t i c e c e r t a i n d i f f e r e n c e s and c e r t a i n similarities between this familiar kind of projection and the kind of geometric proj e c t i o n which is presented. 10-3.
--
Definition: The projection of a point i n t o a plane is the foot of the perpendicular f r o m the p o i n t t o the p l a n e . (BY Theorem 8-10 t h i s perpendicular exists and is unique. )
In the f i g u r e ,
Q is the projection of
---
P
into
E.
Definition: The p r o j e c t i o n of a line into a plane is the set of points which are p r o j e c t i o n s into the p l a n e of the p o i n t s of the l i n e .
In the figure, P* is jection of Q, and so on. line I s a line; and in fact when the l i n e and the plane
the projection of P , Qt is the proIt looks as if the pro jection of the this i s what always happens, except are perpendicular.
-
Theorem 10-7. The projection of a line into a plane is a line, unless t h e l i n e and the plane a r e perpendicular.
Proof:
L be a line n o t perpendicular
Let
to
plane
E.
L lies in E. Then e a c h p o i n t o f L lies In E and is its own projection, (That is, a line through such a p o i n t P, perpendicular to E, I n t e r s e c t s E in P. ) Thus, the projection of L is just L itself, and so is certainly a line. Case 1.
does n o t lie in
E. L e t P be a point of L that i s not in E, l e t PI be the projecttion of P into E, and l e t F be the plane determined by the intersecting lines L and I? and E have point P I In common, ane so, they I n t e r s e c t in a line which we c a l l L1 . ( p o s t u l a t e 8 . ) We want to show t h a t L1 i s t h e p r o j e c t i o n of L. Case 2.
L
EL
To do t h i s we must show two things: (1) If R is a p o i n t of L, then i t s projection is a point of L' This will show that the projection of L lies on L t , but I t will not assure us that the projection of L constitutes all of Lt To show the l a t t e r we must prove
.
.
is any p o i n t of L* t h e r e is a point L whose projection is S t .
(2) If
St
S of
We can prove these two p a r t s of Case
2
as f o l l o w s :
If R = P, then R1 = P t and so R 1 l i e s on Lt . So suppose R is d i f f e r e n t from P. Then EP*' and ^n?a are coplanar, by Theorem 8-8. Since F I s the only plane containing P, R and P I ( ( ~ o s t u l a t e7.1, Rr I s in F . R* is also in E. Therefore R1 is on L', since L t , being the intersection of E and F, contains all points common to E and P. Proof of (1):
-
-
P r o o f of (2): If S Ã is any point of Lt, then the line M through S t perpendicular to E I s coplanar with P P t ( o r coincides with it if S t = PI) and so lies in F . Therefore M i n t e r s e c t s L (why?) at some p o i n t S . S t is the projection of S. This completes the proof of Theorem 10-7.
If a line is perpendicular to a plane its projection i n t o the plane is a single point. The idea of projection c a n be defined more generally, f o r any set of p o i n t s . If A is any s e t of points, then the projection of A Into the plane E is simply the set of all pro- . jections of p o i n t s of A . For example, the projection of a triangle is usually a triangle, although In certain exceptional cases it may be a segment. Q
'
APQRIs ASW. On the right, contains APQRis perpendicular to E, so t h a t of APQR is simply t h e segment ST.
On t h e l e f t , the projection of the plane t h a t
the projection
--
Problem S e t 10-3
1.
Using the kind of projection explained in S e c t i o n 10-3 answer the following:
Is the projection of a p o i n t always a point? Is the projection of a segment always a segment? Can the projection of an angle be a ray? An angle?
A line?
Can the projection of an a c u t e angle be an obtuse angle?
Is the projection of a
right angle always a right angle?
Can t h e length of the projection of a segment be greater than the length o f the segment?
If two segments are congruent will their projections be congruent?
If two lines Intersect can t h e i r projections be two parallel lines?
If two lines do not i n t e r s e c t can t h e i r projections be two i n t e r s e c t i n g lines? If two segments are p a r a l l e l and congruent, will t h e i r projections be congruent?
3.
Given the figure with n o t in plane m, XY the projection of into plane rn, M t h e midpoint of E , and N the p r o j e c t ion o f M, prove N is the mid-point of E.
Front
2 Top View
In mechanical drawing the top view or "plan" of a solid may be considered the projection of the various segments of the s o l i d i n t o a horizontal plane m, as shown in perspective a t the left. The t o p view as it would a c t u a l l y be drawn is shown at the right. (NO attempt is made here to give dimensions to the segments. ) a.
Sketch a f r o n t view of the s o l i d shown above - t h a t is, sketch the r e s u l t of projecting t h e segments of the s o l i d Into any plane parallel to its front face.
-
b.
Sketch the right s i d e view of the solid.
5.
The projection of a tetrahedron (triangular pyramid) into the plane of i t s base may look like the figure at the right. How else may it appear?
6,
Given:
-
BD is t h e projection
U e s ~ p l a n e m~ is a right a n g l e .
Prove: angle.
/
ABD
( ~ i n t : Let l a r to plane
A
B
C
/7/^-1
l a a right
BE m. )
be perpendlcu-
L___________/
*7.
+
Given: A Q Has projection + * AR in plane m. AP is any other ray from A in plane m. (~ofce: / QAR 1s called the m ~ l ethat A$ makes with plane m.) Prove:
m/ QAR
<
m/ QAP ,
( ~ i n t : Let Q1 be t h e projection of Q i n t o m, On AP choose X so t h a t AX = A Q 1 .
-
*8.
-
QQ1, Q f X
Draw
and
OX.)'
If t h e diagonal of a cube is perpendicular to a given plane, sketch the p r o j e c t i o n into t h e plane of all the edges of the cube.
(NO
proof r e q u i r e d . )
Review Problems
/
R-AB-S is an a c u t e dihedral angle w i t h
Suppose
P a point on its
* PX
edge. 4
Can rays and PY be chosen in the t w o faces so that a. b. c.
/ / /
XPY
is a c u t e ?
XPY
is obtuse?
XPY
is r i g h t ?
2.
Planes r and s Intersect Çin TQ. 3 is a p o i n t between T and Q. i s i n r.
m/ TBA m/
FBQ
=
FB
is in s . 90. Is / ABF a
= 40.
plane angle of dihedral
/
TQ? Can
you determine
If SO, s t a t e a theorem to support your conclusion.
m/ ABF?
3.
Planes x and r intersect In %. B is a p o i n t between + K and Q. BA i s ~ nr. "B? is in x. m/ ABK = 90. m/ QBF = 90. Is / FBA a plane angle of dihedral / QK? If your answer is "yes", s t a t e a theorem or d e f i n i t i o n to support your conclusion. If m/ ABF = 80, is r l x ? If r l x , what is m/ ABF?
4.
Indicate whether each of t h e following statements is t r u e in all cases ( A ) , tme in some cases m d false in others (31, or true in g caae (N).
a.
Two lines parallel to the same plane are perpendicular to each other.
b.
If a plane i n t e r s e c t s each of t ~ Ion t e r s e c t i n g planes, the lines of i n t e r s e c t i o n are parallel.
c.
If a line lies in a plane, a perpendicular to the line I s perpendicular to the plane.
If two planes are parallel, any line in one of them is parallel to the other.
If two planes are parallel t o t h e same line they are parallel to each o t h e r . Two lines perpendicular t o the same line at the same point are perpendicular to each other.
If t w o intersecting planes are each perpendicular t o a third plane, t h e i r line of i n t e r s e c t i o n is perpendicular to the t h i r d plane. The projection of a segment is a segment.
The projection of a right angle is a right angle. Congruent segments have congruent projections. Two lines are p a r a l l e l if they are both perpendicular
to the same line. If a plane is perpendicular to each of two lines, t h e two lines are coplanar. If a plane I n t e r s e c t s two o t h e r planes In parallel l i n e s , then the two planes are parallel.
If a plane i n t e r s e c t s the faces of a dihedral angle, t h e i n t e r s e c t i o n is c a l l e d a plane angle of t h e dihedral angle.
5.
Given: F i s the projection of p o i n t A into plane E. BH lies in plane E. / FBH is a right angle.
Prove: angle.
/
ABH
is a right
6.
Given: Planes X, Y and Z a r e parallel as shown, with in 2, and A in X. AC cuts Y in B and c u t s Y In D. A 3 = BC. AC = CE.
Prove:
7.
BD = BA.
Given: R , 2, Y, X are t h e mid-points of the res p e c t i v e s i d e s CB, BA, - AD, DC of the non-planar
- -
quadrilateral Prove: gram.
*8.
RZYX
A
CBAD .
is a p a r a l l e l o -
In the following incomplete s t a t e m e n t it is possible to fill in the solid blanks with "line" or "plane" and t h e dotted blanks w i t h I [ or in eight ways so as to make the completed statement true: Give f i v e of these ways.
1
Given:
ABCD is a p a r a l l e l o - gram. Each of E, BF, XY, DH and are perpendicular t o L, L is in the plane o f parallelogram ABCD. Prove:
AE
+
CG = BF + DH*
Appendix I
A CONVENIENT SHORTHAND There was a t i m e when algebra was all written out in words. In words, you might state an a l g e b r a i c problem in the following way: 11
If you square a c e r t a i n number, add f i v e times the number, and then subtract s i x , the result is zero. What are possibilities for t h i s number?" This problem can be more b r i e f l y s t a t e d in the following form: ' ~ i n dthe r o o t s of the equation 2 + 5x - 6 = 0." The n o t a t i o n of algebra I s a very convenient shorthand. A similar shorthand has been Invented for talking about s e t s . It saves a lot of time and space, once you g e t used to I t , and it is a l l right t o use it In your written work, unless your teacher ' objects Let us start with a picture, and say various t h i n g s about i t first in words and then in shorthand. L
.
2
Here we see a line L, two half-planes H1 and H,,. ways. In Words 1. 2.
PQ
which separates the plane E i n t o Now l e t us say some things in t w o
The segment lie8 InHl. The i n t e r s e c t i o n of and L IS T.
In Shorthand
mC
1. H . 2. 1^ 0 L = T.
-
The shorthand expression PQ C H, is pronounced in exactly the same way as the expression on the left of I t . In general, when we write A C B, t h i s means t h a t the set A lies in the set
;
B. An expression of the type
nB
denotes the intersection of t h e s e t s A and B. The symbol " 0 '' i s pronounced "cap," because I t looks a little like a cap. Notice t h a t the sets PQ and RS do not intersect. If we agree t o write 0 f o r the empty set, then we can express this fact by writing F Q ~ R =S 0 . Similarly, L = O A
-
p^n
and
a
mfl H,
= 0.
Of course, is a s e t which lies In H,. But the point P above I s a member of H,. We write t h i s In shorthand l i k e t h i s P C : H,. T h i s I s pronounced "P belongs to Hi." The union of t w o s e t s A and B is written as A U B. T h i s is pronounced " A cup B." In the same way, we write A U 3 U C f o r t h e union of three s e t s . For example, in the figure above, the plane E I s the union of H,, Hp and L. We can therefore write E = H , U HpU L. Notice t h a t h e r e ( a s everywhere else), a formula i n v o l v i n g the sign "=" means that the things on the left and r i g h t of "=" are t h e same thing. The sign "=I1 i s simply an abbreviation of the word "is", as in t h e expression 2 + 2 = 4, which says t h a t two plus t w o is four. Problem Set -I Consider the s e t s , A, B, C , and so on, d e f i n e d in the foilowIng way:
A I s the set of all doctors. B I s the s e t o f a l l lawyers. C is t h e s e t of a l l tall people.
1
1 1 1 I
D is t h e s e t of all people who c a n play the v i o l i n . E is t h e s e t of all people who make a l o t of money. F is the set of a l l basketball players. Write shorthand expressions f o r t h e following statements: 1. All basketball players are tall. 2.
3. 4.
5. 6. 7. 8. 9.
No d o c t o r is a lawyer. No violinist makes a l o t of money, unless he is tall. No b a s k e t b a l l player I s a violinist, Everyone who is both a d o c t o r and a lawyer can also play the violin. Every basketball lot of money. The man X is a The man Y is a The man Z is a
player who can play the violin makes a tall violinist. prosperous lawyer. tall basketball player.
Appendix I1
POSTULATES OF ADDITION A N D MULTIPLICATION The methods of manipulating real numbers by means o f t h e operation of addition and mu1 tiplication, and the related o p e r a t i o n s of s u b t r a c t i o n and d i v i s i o n , are a l l determined by the following eleven p o s t u l a t e s . I n t h e statement of these postulates and t h e proofs of the following theorems It is to be understood that all the l e t t e r s are real numbers. A-1. (closure under Addition.) x + y is always a r e a l
number.
+ (y + z ) = ( x + + y = y + x.
A-2.
(~ssociativeLaw for ~ddition.) x
A-3. A-4.
(cornmutati v e Law f o r Addition. ) x (Existence of 0.) There is a unique number 0 such
y)+z.
+ 0 = x f o r every x . A-5. (Existence of ~ e g a t i v e s . ) For each x t h e r e is a unique number -x such t h a t x + ( - x ) 0. that
x
Â¥
M-1.
(Closure under ~ulti~lication.)xy
is always a real
number. M-2.
( ~ s s o c i a t i v eLaw f o r ~ultiplication.) x(yz) = (xy)z. M-3. (commutative Law for M u l t i p l i c a t i o n . ) xy = yx. M-4. (~xistenceof 1. ) There is a unique number 1 such t h a t xal == x for every x . M-5. (~xistenceof Reciprocals. ) For each number x o t h e r l such that x--x1 = 1. than 0 there is a unique number 2 ( D i s t r i b u t i v e Law. ) x ( y + z) = xy + xz. D. The following basic theorems will illustrate how these post u l a t e s are used in simple c a s e s . Theorem 11-1. If b = -a,then -b = a. Proof: By A - 5 , b = -a means the same as a + b = 0. By A-3 t h i s is the same as b + a = 0. By A-5, this is the same as a = -b.
Another way of stating this theorem is that
-(-a) = a .
Theorem 11-2. For any a , a - 0 = 0. Proof : a = agl = a ( l + 0) = a - 1 + a.0 = a + a-0 Hence by (A-41, a - 0 = 0. Theorem 11-3. a(-b) = -(ab). Proof:
(M-41
(A-4) 04-4)
ab + a(-b) = a [ b + (-b)] (Dl = a.0 (A-5) = 0 ( ~ h .11-2) Hence by A-4, a ( - b ) = -(ab). A s a special case of this theorem we have a(-1) = - a . D e f i n i t i o n . x - y shall mean x -I- (-y). Note that by this definition a - a = 0. Theorem 11-4. if a + b = c , then a = c b. Proof: If a + b = c , t h e n (a + b) + (-b) = c + (-b) ( a + b ) + (-b) = a + [ b + (-b)] (A-2)
-
a
+
0
(A-51
-
(A-4)
= a
Hence
a = c I- (-b) = c
-
by definition. Theorem 11-5. If ab = 0, then e i t h e r a 0 or b = 0. Proof: To prove the theorem I t will be enough to show that 1 e x i s t s , by If a # 0 then b = 0. So suppose a # 0. Then g b
Â¥Â
M-5.
Therefore,
-a1
1 (ah) = -*o = a
o
( ~ h .A-11-2)
1 a
1 (ab) = (Ña = 1-b
b
(M-2)
also,
Therefore
b = 0.
04-51
(Cancellation Law.)
Theorem 11-6. then
b = c. P r o o f : If
ab = ac
then
ab
-
If
ab
= ac
and
a
#
0
ac = 0 , By Theorem A-11-3
-
this i s the same as ab + a ( - c ) = 0, or, by D, a s a ( b c ) = 0. Since a # 0 we g e t , by applying Theorem 11-5, that b - c = 0. Hence b = c . These are j u s t a few examples of t h e use of the postu-lates in proving basic algebraic theorems. Ordinarily we don't use the postulates d i r e c t l y but make use of such properties as those stated In Theorems 11-4 and 11-6 in our algebraic work.
Problem Set I1 -
1.
7
Prove each of the following theorems.
(-a) (-b) = ab. b. a(b-c) = ab - ac. c . If a - b = c , then a = b + c . d. (a + b) (c + d) = ac + ad + be + bd. ( l i i n t : As a first s t e p apply D, regarding ( a + b) as a single number. ) 2. Given the d e f i n i t i o n s : 2 a.
x
= X'X,
prove t h a t 3.
4.
3
.
( a + b12 = a2 + 2ab + b 2 2 Prove: (a + b) (a-b) = a - b
.
-1 Definition: = ab Prove each of the following: a, b. c.
-y1 (ab)-1 = a -a. -c= - ac b d bda ac
x==.
-1
a.
( - a ) 1 = -(a ). a -a a e. -5 = = -5
.
f.
g.
a + c
b
to
a
c
5 + g =
b
ad
+
bd
be
.
Appendix 111 NATIONAL AND IRRATIONAL NUMBERS
111-1. ----How t o Show That a Number & Rational. By d e f i n i t i o n a number i s r a t i o n a l i f i t i s t h e r a t i o of two integers. Therefore, it we want to prove that a number x is rational, we have to produce two i n t e g e r s p and q, such that P = x . Here are some examples: q (1) The number* x = + Is rational, because
-
x = ¥^ where p = 13 amd q = 14. Q (2) The number x = 1.23 is rational, because 1.23 = 123 100' which I s the ratio o f t h e two integers 123 and 100. (3) If t h e number x I s rational, then so is the number 2x. h hat is, twice a rational number is always rational.) For if Therefore
-
where
p
and
q
are Integers, then
where the numerator 2p ( 4 ) If the number 2 x + 3. For if
2x - q9 , and t h e denominator q are both integers. x I s rational, then so is the number
then
where the numerator and denominator are b o t h Integers. . 5 ) If x is a rational number, then so I s xz + x. x = -P , q
then
where t h e numerator and denominator are integers.
For if
--
Problem S e t 111-1
3. 4.
Show that .£35 Is a r a t i o n a l number. Show that -7 + 5 is ~ational. 7 Show that if x is a rational number, then so I s Show that If x is rational, then so is 2x 7.
5. 6.
Show that 7 + - is rational. 17 Show t h a t the sum of two rational numbers is a
1. 2.
-
x - 5.
rational number. 7. Show that (-)17 ( 23 ) I s rational. T7two ratlonal numbers is a 8. Show that t h e product of rational number. 9. show t h a t 23 1s ~ a t ~ o n a l . 17 +-? two rational numbers is a 10. Show t h a t the qu&ient of rational number, as long as t h e d i v i s o r i s not zero. 11. Given t h a t i/^T is irrational, show t h a t ^t 1s also I r r a t i o n a l . ( H i n t : This problem is a lot easier, now that you understand about Indirect proofs .) 12. Given that T is irrational, show that Is also 5
13. 14.
15.
16. 17.
irrational. Show that t h e reciprocal of every rational number d i f ferent from zero is rational. Show that the reciprocal of every irrational number d i f ferent from zero Is irrational. Is it true t h a t the sum of a rational number and an irrational number is always irrational? Why or why n o t ? Is it true t h a t the sum of t w o Irrational numbers is always Irrational? Why or why not? How about the product o f a rational number and an irrational number?
111-2. Some E w n p l e s & Irrational N m b e ~ s . I n the previous section, we proved t h a t under certain condit i o n s a number must be rational. In some of the problems, you showed t h a t starting with an Irrational number we could get more irrational numbers in various ways. In all this we l e f t one very important question unsettled: are there any Irrational numbers? We shall settle t h i s question by showing t h a t a p a r t i c u l a r number, namely-^/^ cannot be expressed as the r a t i o of any two i n t e g e r s . To prove this, we f i r s t need to establish some of the facts about squares of odd and even I n t e g e r s . Every integer I s e i t h e r even or odd. If n is even, t h e n n is twice some integer k, and we can write n = 2k. If n is odd, then when we d i v i d e by 2 we g e t a quotient k and a remainder 1, so that
--
Therefore, we can write n = 2k + 1. These are t h e typical formulas for even numbers respectively. For example, 6 = 2-3 n = 6 , k = 7 = 2-3 + 1 n=7, k = 8 = 2.4 n a 8 , k =
and odd numbers
,
3 3 4
The f o l l o w i n g theorem is easy t o prove: Theorem 111-1. The square of every odd number Is odd. Proof: If n I s oda, then we can write n = 2k + 1, where k is an I n t e g e r . Squaring both sides, we get n 2 = ( ~ k +) 2' ~21t + 1 2 = 4k + 4k 4- 1. The right-hand s i d e must be odd, because it I s written I n the form 2'[2k2 + 2k] + 1; I s odd, that l a , It Is twice an integer, plus 1. Therefore, n which was to be proved,
and so on.
I
I
From Theorem 111-1 we can q u i c k l y g e t another theorem: Theorem 111-2. If n2 I s even, then n is even. Proof: If n were odd, then n2 would be odd, which is f a l s e . Therefore n I s even. Notice t h a t t h i s is an i n d i r e c t proof. We are now ready to begin the proof of Theorem 111 -3. f i is i r r a t i o n a l . Proof: The proof will be indirect. We begin by making the assumption t h a t i/^T I s rational. We w i l l show that this leads to
a contradiction. Step 1. Supposing t h a t can be expressed as
./2"
is rational, it follows that
2
^
Â¥wher the fraction is In lowest terms. q The reason is that if i/T can be expressed aa a fraction a t all, then we can reduce the fraction to lowest terms by d i v i d i n g out any common f a c t o r s of the numerator and denominator. We therefore have in lowest terms.
This gives
2 = -P* 0
)
which in turn gives
Step 2. p2 is even. Because p2 is twice an Integer. Step 3 . p I s even. By Theorem 111-2. We therefore s e t p = 2k. Substituting In the formula at the end of Step 1, we get 2 2 (2k) = 2q , which means t h a t
za2
.
q 2 = 2 k2
.
4k2 =
Therefore
s t e p ^. Because
even. Is t w i c e an I n t e g e r .
S I
Step 5. q is even. By Theorem 111-2. We started by assuming t h a t */^ was r a t i o n a l . From this we g o t #/T= 2 in lowest terms. From t h i s we have proved t h a t p q' and q were both even. Therefore was not in lowest terms, q a f t e r a l l . This contradiction shows that our i n i t i a l assumption must have been wrong, that Is, i/^T must not be rational.
--
Problem Set 111-2
These problems are harder than most of the problems in the text.
6 is
irrational, so as to g e t a proof that 1/3" is Irrational. (!lint : S t a r t with t h e f a c t t h a t every integer has one of t h e forms n = 3k n = 3 k + 1 n = 3k + 2, and then prove a theorem corresponding to Theorem 111-2.) 2, Obviously nobody can prove that f i is irrational, because 6a: 2. If you try to "prove" this by adapting t h e proof f o r ^/?, at what p o i n t does the "proof" break down? 3 . Show that 'fi I s Irrational. Actually, t h e square r o o t of an i n t e g e r is e i t h e r another integer or an irrational number; t h a t is, */n' e i t h e r "comes o u t very e v e n or "comes o u t v e r y uneven.'' The proof of this f a c t , however, requires more mathematical technique than we now have at our disposal. Problems like this are solved In a branch of mathe n i a t l c s called the Theory of Numbers. 1.
Adapt the proof that
Appendix IV SQUARES AND SQUARE ROOTS
Everybody knows what I t means to square a number: you multiply the number by itself. The f a c t s about square roots, however, are considerably trickier, and the language in which most people talk about them is very confusing. Here we will try to s t a t e the f a c t s and p o i n t out t h e p i t - f a l l s . To say that x is a square r o o t of a -means t h a t 2
x
= a.
For example, a square root of 4, a square r o o t of 9, a square root of 4, a square r o o t of 9 , and so on. You may wonder why we d i d not abbreviate these statements by using radical signs. The reason (as w e s h a l l soon see) is t h a t radical s i g n s mean something s l i g h t l y different. The following l a a fundamental fact about the real number system: Is 3 is -2 is -3 i s 2
I
Every p o s i t i v e number has exactly one p o s i t i v e square r o o t .
For example, 22 = 4, and no other positive number I s a root of the equation x2 = 4. 4 = 16, and no other p o s i t i v e number I s a root of the equation x2 = 16. And so on. Of course, if x I s a square r o o t of a, t h e n so is -x, 2 2 because ( - x ) = x . Therefore every positive number has exactly two square roots, one positive and the o t h e r n e g a t i v e . The meaning of the radical sign I s defined t h i s way:
-
If a is p o s i t i v e , t h e n f i square root of a.
denotes the p o s i t i v e
We provide further
that -/O = 0.
For example,
^ 8 ^
= 2, = 3, =
4,
and so on. To i n d i c a t e the other square root negative one we simply put a minus sign In sign. For example: 4 has two square r o o t s , 2 3 has t w o square roots, ,/3" 7 has two square r o o t s , ./T The following two statements look a l i k e , different: (1) x is a square root of
--
(2) x
--
t h a t is, the front of the radical
and and and
-2.
- a.
- È/T but i n f a c t they are
a.
=/S-.
2 The first statement means merely t h a t x = a. The second statement means not only that x2 = a , but a l s o that x > 0. Therefore the second statement I s not simply a short-hand form of the first. L e t us now investigate the expression-i/x, where x is not equal to zero. There are t w o possibilities: I. If x > 0, then x is the positive square root of x2, and we can write
If x < 0, then x is the negative square root of x2, and I t is -x that is the p o s i t i v e square root of x2 . Therefore, f o r x < 0, we have 11.
^s
-X.
looks so appealing t h a t I t seems almost l i k e a law of nature. In f a c t , however, this equation holds true only half of the time: it I s always true when x 2 O? and I t is never true when x < 0.
----
Fitting together cases without exception we have
11, we see t h a t f o r every
I and
J?-
1x1To see tills, you should check it against the definition of S e c t i o n 2-3.
1 x 1,
x
in
--
Problem S e t IV Which of the following statements
1. 2. 3.
4.
^9
are true? Why or why n o t ?
= 3.
-3.
= =
4-
= =
2 5. 5.
-
(~pproxirnately.) (~pproximat ely )
1.414. r 2 = 1.414.
5.
.
6. y e 5 For what values of the unknowns (if any) do the following equations hold true? Why? 7 . .^(-= a 1.
-
9 10. 11. 12. 13. 14.
,/X= -1x - 11. ./(X = (x 312 2 4-
J +
= --(x + 3 )
./(x ,/(xm = + =
3)" (x
+ 3I2l
l(x
31~1.
Appendix V HOW TO DRAW FIGURES IN 3-SPACE
.
Simple Drawae. A course in mechanical drawing is concerned w i t h precise representatlon of physical objects seen from d i f f e r e n t positions in space. In geometry we are concerned with drawing only to the ext e n t t h a t we use sketches t o help us do mathematical thinking. There I s no one correct way to draw pictures I n geometry, but t h e r e are some t e c h n i q u e s h e l p f u l enough t o be In rather general u s e . Here, f o r example, I s a technically correct drawing of an o r d i n a r y pyramid, f o r a person can argue t h a t he is looking at t h e pyramid from d i r e c t l y above. Bat c a r e f u l ruler drawing is n o t as helpful a s this very crude free-hand sketch. The first drawing does not suggest 3-space; the second one does. The f i r s t part of t h i s discussion offers suggestions f o r simple ways to draw 3-space f i g u r e s . The second part i n t r o d u c e s the more elaborate technique o f drawing from perspective. The difference between the two approaches is suggested by these t w o drawings of a rectangular box.
V-1
€
/q)
In t h e f i r s t drawing the base is shown by an easy-to-draw p a r a l l elogram. In t h e second drawing, the f r o n t base edge and t h e back base edge are parallel, but the back base edge is drawn s h o r t e r under the belief t h a t the shorter length will suggest "more remote".
No matter how a rectangular box is drawn, some s a c r i f i c e s must be made. All angles of a rectangular s o l i d are right angles, but in each of the drawings shown above two-thirds of the angles do not come c l o s e t o I n d i c a t i n g ninety degrees when measured w i t h a protractor. We are willing to give up the drawing of r i g h t angles t h a t look like r i g h t angles in order t h a t we make the f i g u r e as a whole more suggestive. You already know t h a t a plane is generally pictured by a parallelogram. It seems reasonable to draw a horizontal plane In e i t h e r of t h e ways shown, and to draw a v e r t i c a l plane like t h i s .
L 3n
0
If we want t o indicate two parallel planes, however, we can not be e f f e c t i v e if we just draw any two " h o r i z o n t a l " planes. Notice how t h e drawing t o t h e right below Improves upon the one t o t h e l e f t . Perhaps you p r e f e r s t i l l a n o t h e r kind of drawing.
Various d e v i c e s are used t o Indicate t h a t one part of a f i g ure passes behind a n o t h e r p a r t . Sometimes a h i d d e n part Is simply omitted, sometimes it is indicated by dotted l i n e s . Thus a line piercing a plane may be drawn i n e i t h e r of the t w o ways:
Two intersecting planes are I l l u s t r a t e d by each of these drawings.
The second I s b e t t e r t h a n the first because t h e line o f
Inter-
section is shown and parts concealed from view are dotted. The t h i r d and f o u r t h drawings are better y e t because t h e l i n e of i n tersection I s v i s u a l l y t i e d i n with plane P as well as plane Q by t h e use of parallel l i n e s I n the drawing. Here is a drawing which has the advantage A o f simplicity and t h e disadvantage of suggesting one plane and one half-plane. In any case a line of i n t e r s e c t i o n is a p a r t i c u l a r l y important part of a figure. Suppose t h a t we wish to draw two Intersecting planes each perpendicular t o a t h i r d plane. An effective procedure is shown by this step-by-step development.
Notice how the l a s t two planes drawn a r e b u i l t on the l i n e of I n t e r s e c t i o n . A complete drawing showing a l l the hidden lines I s just too Involved t o handle pleasantly. The picture below is much more suggestive. 1
I
1
A dime, from d i f f e r e n t angles, l o o k s l i k e this:
0 Qa Neither t h e f i r s t nor the last is a good p i c t u ~ eof a c i r c l e fn 3-space. E i t h e r of t h e o t h e ~ sis satisfactory. The thinner oval is perhaps better to use to represent the base of a cone,
C e ~ t a i n l ynobody should expect us to i n t e r p r e t the f i g u r e shown
A few addL tional drawings, w i t h verbal d e s c r i p t i o n s , ape shown. A line parallel t o a p l a n e .
A cylinder cut by a plane
parallel to t h e base,
A c y l i n d e r c u t by a plane not parallel to t h e base.
A pyrmld cut by a plane parallel t o t h e base.
It is important to remember t h a t a
is not an end in I t s e l f but simply an aid to our understanding of the geometrical sttuation. We should choose the kind of p i c t u r e that wlll serve us best f o r this purpose, and one personts chofce may be d i f f e r e n t from another. drawing
Perspective. The rays a ? b ? c, d, e , f In t h e left-hand f i g u ~ ebelow suggest coplanar l i n e s intersecting at V; the corresponding ray6 in the right-hmd Xigure suggest parallel lines in a three-dimensional drawing. Thlnk of a railroad t r a c k and telephone poles as you look at t h e rig11-k-hand figure
V-2.
The right -hand figure suggests certain principles which are useful in making perspective drawings. 1) A s e t of p a ~ a l l e l line^ which recede from t h e viewer are d~awnas concurrent m y s ; f o example, ~ rays a, b, c, d, e , f. The p o i n t , on the drawing? where the rays met is h o r n as the " vanishing point" 2) Congruent segments are dram smaller when they are farther from t h e vLewer. (Find examples 5n t h e drawing. )
.
3 ) Parallel lines which are perpendicular t o t h e l i n e of s i g h t of t h e viewer are shown as parallel l i n e s in t h e dra&.ng. (FLnd examples An t h e drawing. A pepson does n o t need much a ~ t i s t l cability to make use of these t h m e principles. The steps t o follow in sketching a rectangular s o l i d are ~ h o w nbelow.
m
Draw the f r o n t face as a rectangle* Select a vanishing point and draw segments *om J t to t h e v e r t i c e s . Omit segments t h a t cannot be seen. Draw edges p a r a l l e l t o t h o s e of
the front f a c e . Finally erase l i n e s of p e r s p e c t i v e .
Under this technique a single horizontal plane can be dram a s t h e t o p face of the s o l i d shown above.
A single vertical plane
can be pepresented by the f ~ o n tface or
the right-hand face of the solid. After t h i s brLef account o f two approaches t o t h e drawing of figures in 3-space we should once again recognize the f a c t t h a t there is no one correct way to picture geometr5c ideas. However, t h e more "rezl" we want our p i c t u r e to appear, t h e more a t t e n t i o n we s h o u l d pay to perspective. Such an artist as k o n a ~ d o& V l n c i paid great attention to perspective. Most of us find this done f o p us when we use o r d i n a r y cameras. See some books o n drawlng or l o o k up "perspectivett in an encyclopedia i f you a r e i n t e r e s t e d T n a d e t a l l e d treatment.
Appendix VI PROOFS OF THEOREMS ON PERFENDICUURITY
In Section 8-3 two themems are s t a t e d , which, between them, cover all cases of existence and uniqueness involved In the perpendicularity of a l i n e and a plane. As s t a t e d t h e r e , e i g h t separate items must be proved to establish the p ~ o o f 8of these two theorems. Here we will s t a t e these items and prove those which have not already been proved. We f i r s t r e s t a t e the two theorems.
u.Through
a glven point there is one and only one plane perpendicular to a given line.
Theo~ern
-
-
Theorem 8-10. Through a given p o i n t there is one and only one line perpendicular to a given plane.
-
We now consider the eight p r o o f s , in a systematic o r d e r .
Read the statements carefully, for there are only slight differences in t h e i r wording: the presence or absence of' a "not", the substitution of "mo8t1' f o r "leastt', o r t h e interchange of "line"
and l1p1ane'l.
-
D
t
I
Theorem V I - 1 . Through a given point on a given l i n e t h e r e is a t least one plane pwpendicular to the line. This is meorem 0-4, which i s proved i n t h e t e x t .
Theorem VI-2, Through a given p o i n t on a glven l i n e there is a t m o s t one plane perpendicular to the l i n e .
This is Theorem 8-6, which I s proved in the t e x t .
meorern VI-3. Through a given p o i n t not on a given l i n e there is at least one plane perpendicular to the given lineGiven:
Line
To prove:
P n o t on L.
L and point
There is a plane
There 1s a line
(meorem 6 - 4 ) . l i e I n the plane
E
through
P,
wlth
F
1 L,
F perpendicular t o L i n t e r s e c t a t Q, [Theorem 3-41.
through M and
M ~t
I3
L and
There i~ a point R ( ~ i g u r e2 1 not in F ( postulate 5b) , L e t G be the plane conhlning L and R (meorem 3-31.
In G
there 1s a llne N
perpendicular to
L at Q
( n e o ~ r n6-11.
L e t E be the plane containing M by Theorem 0-3,
and
N.
Then E
1 I,
-
Theorem VI-4. Through a glven point not on a given llne there 1s at most one plane perpendicular t o the given line. Proof: Suppose that there are two planes perpendicular t o line L and each containing
El and P.
E2, each 1f El
and E2 intemect L in the same p o i n t Q, we have two planes perpendicular to L at Q, m d t h l s contradicts Theorem VI-2.
Ep intersect L In d i s t i n c t @ points A and B , then PA and PB are d i s t i n c t l i n e s through P perpendicular to L, contradicting Theorem 6-4. E i t h e r way, we g e t a contradiction, and so we cannot have two planes through P perpendicular to L.
On the other hand, if El and <Ñ
T h i s finishes the proof of Theorem 8-9.
The next four
theorems, which read like the p r e v i o u s f o u r w i t h "line" and \plane" Interchanged, will prove Theorem 8-10.
-
Theorem VI-5. Through a given point in a given plane there is at l e a s t one line perpendicular to the plane.
A
Proof: L e t P be a p o i n t in plane E. By Postulate 5a there is another p o i n t Q i n E. Let plane F be perpendicular <-à to PQ at P (?Theorem VI-1) Since F intersects E ( a t P) t h e i r i n t e r s e c t i o n i s a l i n e M, by Postulate 8. L e t L be a line in F, perpendicular to M (Theorem 6-1) Since F and L lies in F and contains P, we have, from the definition of a line perpendicular to a plane, that <6Ñ L PQ. A l s o , from above, L _^ M. Hence L E, by Theorem 8-4.
.
1 z,
.
1
1
-
Theorem VI-6. Through a given p o i n t in a given plane there is a t most one line perpendicular to the given plane. Proof: Suppose L-, and L2 are distinct lines, each perpendicular to plane E at point P. Li and L2 determine a plane F h he or em 3-4) which i n t e r s e c t s E in a line L. In F, we then have two perpendiculars t o L a t the same p o i n t P i contradicting Theorem 6-1.
Theorem VI-7. Through a glven pomt not a glven plane there is at least one line perpendicular t o the given plane.
Proof: L e t P be a point not in plane E. Let A be any p o i n t of E, and M a line through A perpendicular to E (!Theorem VI-5). If M
If M
contains
P I t is the desired perpendicular. does n o t contain P l e t F be the plane containing
M and P h he or ern 3-31, and N the line of Intersection of F and E. In F l e t 3 be the foot of a perpendicular f r o m F to N heo or ern 6-4). L e t line L be perpendicular to E at 3 (Theorem VI-5) By Theorem 8-8, L and M are coplanar, and hence, L lies in P since M and B detennlne F. In F, L N, s i n c e L E and N lies in E, Since by Theorem 6-1 t h e r e is only one line in F perpendicular to N a t w 3, L and BP must coincide. That l a , L contains P and so is the desired perpendicular.
.
1
1
Theorem VI-8. Through a given point not in a given plane there I s at most one line perpendicular to the given plane. The proof is word f o r word the same as that of Theorem VI-6, except f o r t h e replacement of "at point P" by "from p o i n t P" and of h he or em 6-1" by "!Theorem 6-3".
The -
Meaning and
Use of Symbols
General. =
can be read as "A equals 3 , "A Is equal to B m j "A equal B" (as In "Let A = B") , and possibly other ways to fit the structure of the sentence In which the symbol appears. However, we should not use t h e symbol, , I n such forms as "A and B are ="; its proper use is between t w o expressions. If two expressions are connected n= I t ,I t is to be understood that these two expressions by stand for the same mathematical e n t i t y , In our case either a real number or a point set. " ~ o tequal ton. A # B means that A and B do n o t represent the same entity. The same variations and cautions apply to the use of # as to the use of A = B
-
/.
-
=.
Algebraic.
+, -,-, +.
These familiar algebraic symbols for operating with real numbers need no comment. The basic postulates
<,
about them are presented in Appendix 11. >, 2. L i k e =, these can be read in various ways In sentences, and A < B may stand for the underlined part of " ~ f A I s less than B", "Let A be less than B", "A less than B implies ", e t c . Similarly f o r t h e other three symbols, read "greater than", "less than or equal to", "greater than or equal to". These Inequalities apply only t o real numbers. T h e i r properties are mentioned b r i e f l y in Section 2-2, and In more detail I n Section
<,
7-2.
6, [A[.
n
Square root of An and "absolute value of A". Discussed in Sections 2-2 and 2-3 and Appendix IV.
Geometric. P o i n t S e t s . A s i n g l e l e t t e r may stand f o r any suitably described p o i n t s e t . Thus we may speak of a p o i n t P, a l i n e m, a half-plane H, a c i r c l e C, an angle x , a segment b, e t c . ^¤ The line c o n t a i n i n g the two p o i n t s A and B ( P . 30). The segment having A and 3 as end-points ( P . 45). "51?. The ray w i t h A as i t s e n d - p o i n t and c o n t a i n i n g p o i n t B (P. 45). /ABC. The angle having B a s vertex and B? and 57 as sides (P. 71). AABC.The t r i a n g l e h a v i n g A, B and C as vertices (P. 72). / A-BC-D. The dihedral angle having ^SS as edge and w i t h sides c o n t a i n i n g A and D (P. 299). Real Numbers.
7
The p o s i t i v e number which is the distance between the two p o i n t s A and B, and also the length of the segment (P. 3 4 ) . ~[ABC. The real number between 0 and 180 which I s the degree measure of /ABC (P. 80). Area R. The p o s i t i v e number which is the area of the polygonal region R (P. 320). AB.
Relations.
.
f\*
Congruence. A $ B I s read "A Is congruent to B", but w i t h the same possible variations and r e s t r i c t i o n s as A = B. In the t e x t A and 33 may be two (not necessarily d i f f e r e n t ) segments ( P. log), angles ( P l o g ) , or triangles (P. 111). P e r p e n d i c u l a r . A _^ B is read "A I s perpendicular to B", with t h e same comment as for A and B may be e i t h e r two 1ines ( P 861, two planes (P. 301 ) , or a line and a plane (P. 219).
.
(
.
s.
11
.
A 1 1 B Is read "A is parallel t o B", with the same comment as f o r 2. A and 3 may be either two l i n e s
Parallel.
(P. 241), two p l a n e s (P. 291) or a l i n e and a plane (P. 291).
L i s t o f Postulates
(P. 30) Given any t w o d i f f e r e n t points, there is exactly one l i n e which c o n t a i n s both o f them. Postulate 2. ( P . 3b) h he D i s t a n c e postulate .) To every pair of different points there corresponds a unique p o s i t i v e number. Postulate 3. (P. 36) (The Ruler postulate .) The p o i n t s Postulate 1.
of a l i n e can be p l a c e d in correspondence w i t h the real numbers in such a way that (1) To every p o i n t of the l i n e there corresponds exactly one real number, (2) To every real number there corresponds exactly one p o i n t of the l i n e , and ( 3 ) The distance between two points Is t h e absolute value o f t h e difference of t h e corresponding numbers. Postulate 4. (F. 40) h he Ruler Placement ~ostulate.) Given two p o i n t s P and Q of a l i n e , the coordinate system can be chosen in such a way t h a t the coordinate of P i s zero and the coordinate of Q I s p o s i t i v e . Postulate 5. (P. 54) ( a ) E v e r y p l a n e confcalns at least three non-colllnear points. (b) Space c o n t a i n s a t least four non-coplanar points Postulate 6. (P. 56) if two p o i n t s l i e In a plane, t h e n the line containing t h e s e p o i n t s l i e s In t h e same p l a c e .
.
Postulate 7 - (P. 57) Any three points l i e In at least one p l a n e , and any three non-collinear points lie In e x a c t l y one plane. More b r i e f l y , any three p o i n t s are coplanar, and any three non-collinear p o i n t s determine a plane. Postulate 8. ( P. 58) If two d i f f e r e n t planes intersect, then t h e i r i n t e r s e c t i o n i s a l i n e . Postulate 9. (P. 64) h he Plane Separation ~ostulate.) Given a l i n e and a plane containing I t , the p o i n t s of the plane t h a t do not lie on the line fonn two sets such t h a t (1) each of the s e t s is convex and (2) if P I s I n one s e t and Q I s in the o t h e r then the segment PQ i n t e r s e c t s t h e l i n e .
Postulate 10. (P. 66) he Space Separation p o s t u l a t e . ) The p o i n t s of space t h a t do n o t lie I n a g i v e n p l a n e form two s e t s such t h a t (1) e a c h o f t h e s e t s is convex and (2) if P is in one s e t and Q is in t h e o t h e r , t h e n t h e segrnent PQ intersects the plane. Postulate 11. ( P . 80) h he Angle Measurement Postulate.) To every angle /BAGthere corresponds a real number between 0 and 180. P o s t u l a t e 12. ( P . 81) (The Angle Construction p o s t u l a t e .) L e t 7^ be a ray on the edge of the half-plane H. For every number r between 0 and 180 t h e r e is exactly one ray 'SP, w i t h P I n H, such t h a t ~ / P A B= r. Postulate 13. (P. 81) h he Angle Addition P o s t u l a t e . ) If D is a point in the Interior of /BAG,then ~DA = C + ~/DAC. P o s t u l a t e 14. (P. 82) he Supplement ~ostulate.) If t w o angles form a l i n e a r p a i r , t h e n t h e y are supplementary. P o s t u l a t e 15. ( P . 115) h he S.A.S. ~ o s t u l a t e . ) Given a correspondence between two t r i a n g l e s (or between a triangle and itself). If two sides and t h e i n c l u d e d angle of t h e first triangle a r e congruent t o t h e corresponding p a r t s of t h e second triangle, t h e n the correspondence is a congruence. P o s t u l a t e 16. (P 252) h he Parallel Postulate ) Through a given e x t e r n a l p o i n t t h e r e is at most one l i n e parallel to a g i v e n line. P o s t u l a t e 17. (P. 320) To every polygonal region there corresponds a unique positive number. Postulate 18. (p. 320) 1C two t r i a n g l e s are congruent, then the triangular regions have the same area. P o s t u l a t e 19. (P. 320) Suppose that the region R is the union of two regions Ri and R2 Suppose t h a t Ri and R p intersect at most in a finite number of segments and p o i n t s . Then t h e area of R is t h e sum of the areas of Ri and R p . P o s t u l a t e 20. (P. 322) The area of a rectangle is the product of t h e length of i t s base and t h e l e n g t h of its a l t i t u d e .
-
AD
.
.
.
Postulate 21
. (P . 546)
The volume of a rectangular parallelepiped is the product of the altitude and the area of t h e base. P o s t u l a t e 22. (P 548) (Cavalier11 s Principle. } Given two s o l i d s and a p l a n e . If for every plane which intersects t h e s o l i d s and is parallel to the given plane the two intersections have equal areas, then the two s o l i d s have the same volume.
.
L i s t of Theorems and Corollaries -Theorem 2-1. (P. 42) L e t A, B, C be three points of a line, with coordinates x, y, z . If x < y < z, then 3 is between A and C . Theorem 2-2. (P. 43) Of any three d i f f e r e n t points on the same line, one is between t h e o t h e r two.
(P. 44) Of three d i f f e r e n t points on the same l i n e , only one is between the o t h e r two. Theorem 2-3.
Theorem 2-4. (P. 46) (The Point Plotting he or ern.) Let AB be a ray., and let x be a positive number. Then there I s Ñ exactly one p o i n t P of AB such that AP = x.
+?
Theorem 2-5.
(P. 47)
Every segment has exactly one mid-
( P . 55)
Two d i f f e r e n t lines I n t e r s e c t i n at
point.
Theorem 3-1. most one p o i n t .
Theorem 3-2. (P. 56) fa l i n e intersects a plane not containing It, then the Intersection is a single p o i n t .
Theorem 3-3. (P. 57) Given a line and a p o i n t n o t on the line, there is exactly one plane containing both of them.
Theorem 3-4. (P. 58) Given t w o Intersecting lines, there is exactly one plane containing t h e m .
.
Theorem 4-1. (P 87) both of them are a c u t e .
If two angles are complementary, then
Theorem 4-2.
(P. 87)
Every angle is congruent t o i t s e l f .
Theorem 4-3.
(P. 87)
Any two right angles
are congruent.
Theorem 4-4. (P. 87) If two angles are both congruent and supplementary, then each of them is a right angle. Theorem 4-5. congruent.
(P. 87)
Supplements of congruent angles are
Theorem 4-6. congruent
.
Theorem 4-7.
( P . 88)
~ohplementsof congruent angles are
( P . 88)
Vertical angles are congruent.
Theorem 4-8. (P. 89) If two Intersecting l i n e s form one r i g h t angle, then they fo m four right angles. Theorem 5-1.
(P. 109)
very
segment is congruent to i t s e l f .
Theorem 5-2. (P. 127) If two sides of a triangle are congruent, then the angles opposite these sides are congruent. Corollary 5-2-1. equiangular.
Theorem 5-3.
.
(P 128) Every equilateral triangle is
(P. 129) Every angle has exactly one b i s e c t o r .
Theorem 5-4. ( P . 132) (me A .S .A. Theorem.) Given a correspondence between two triangles (or between a triangle and itself) If t w o angles and the included side of the first triangle are congruent to the corresponding parts of the second triangle, then the correspondence is a congruence.
.
Theorem 5-5. (P. 133) If two angles of a triangle are congruent, the sides opposite these angles are congruent. Corollary 5-5-1. equilateral.
(P. 133) An equiangular triangle is
Theorem 5-6. (P. 137) h he S .S .S. Theorem.) Given a correspondence between two triangles (or between a t r i a n g l e and itself.) If a l l three pairs of corresponding sides are congruent, then the correspondence is a congruence.
Theorem 6-1. (P. 167) In a given plane, through a given point of a given line of the plane, there passes one and only one line perpendicular to the given line. Theorem 6-2. (P. 169) The perpendicular bisector of a segment, In a plane, is the s e t of a l l p o i n t s of the plane that are e q u i d i s t a n t from the end-points of the segment.
.
Theorem 6-3. (P. 171 ) Through a given external point there is at most one line perpendicular to a given line.
(P. 172)
Corollary 6-3-1. can be a right angle.
A t most one angle of
a triangle
Theorem 6-4.
(P. 172) Through a given external point there is at l e a a t one line perpendicular to a given l i n e . Theorem 6-5. line L, then M that contains C .
(P. 183) and A
.
B
/
and C on a are on the same s i d e of any other line
Theorem 6 - 6 . (P 183,) is any p o i n t not on line ABC
.
If M
If M <Ñ
AC,
Is between
A
is between A and C , and then M I s in the i n t e r i o r of
Theorem 7-1. (P. 193) he Exterior Angle Theorem.) An exterior angle of a triangle I s larger than either remote I n t e r i o r angle. Corollary 7-1-1. (P-196) If a triangle has a r i g h t angle, then the other two angles are acute.
Theorem 7-2. (P. 1 9 7 ) (The S . A . A . he or em.) Given a correspondence between t w o triangles. If two angles and a side opposite one of them in one triangle are congruent to the corresponding parts of the second triangle, then the correspondence i s a congruence.
Theorem 7-3. (P. Given a correspondence hypotenuse and one leg corresponding parts of i s a congruence.
198 )
h he
-
Hypotenuse Leg Theorem.) between two r i g h t triangles. If the of one triangle are congruent to the the second triangle, then the correspondence
Theorem 7-4. (P. 200) if two sides of a t r i a n g l e are n o t congruent, then the angles opposite these two s i d e s are n o t congruent, and the larger angle I s opposite the longer aide.
Theorem 7-5. (P. 201) If two angles of a triangle are not congruent, then the sides o p p o s i t e them are not congruent, and t h e l o n g e r s i d e is opposite the larger angle.
1
Theorem 7-6. (P. 206) The shortest segment J o l n l n g a point to a l i n e is the perpendicular segment. Theorem 7-7. (P. 206) (The Triangle Inequality.) The sum of the lengths of any two sides of a triangle is greater than the of the third side. Theorem 7-8. (P. 210) If two sides of one triangle are congruent respectively to two sides of a second triangle, and the included angle of the first triangle I s larger than the included angle of the second, then the opposite slde of the first triangle Is longer than the opposite side of t h e second. Theorem 7-9" [P. 211) If two s i d e s of one triangle are congruent respectively to two sides of a second triangle, and the t h i r d s i d e of the first triangle I s longer than the t h i r d s i d e of the second, then the included angle of the f i r s t triangle is larger than the included angle of t h e second.
Theorem 8-1. (P. 222) If each of two points of a line I s equidistant f r o m two given p o i n t s , then every point of the line is equidistant f r o m the given points. Theorem 8-2. (P. 225) If each o f three non-colltnear points of a plane is equidistant from two points, then every point of the plane is equidistant f r o m these two points.
Theorem 8-3. (P. 226) If a l i n e l a perpendicular to each of two intersecting lines at their point of intersection, then I t Is perpendicular to the plane of these lines.
Theorem 8-4. (P. 230) Through a given point on a given line there passes a plane perpendicular to the line. Theorem 8-5. (P. 231) If a line and a plane are perpendicular, then the plane contains every line perpendicular to the given line at I t s point of Intersection with the given plane.
Theorem 8-6. (P. 233 Through a given point on a given line there I s at most one plane perpendicular to the line.
Theorem 8-7. (P. 232) The perpendicular bisecting plane of a segment I s the s e t of a l l points equidistant f r o m the endpoints of the segment.
Theorem 8-8.
(P. 234) Two lines perpendicular to the
same
plane are coplanar. Theorem 8-9, (P. 235 ) Through a given point there passes one and only one plane perpendicular to a e v e n line.
.
-
Theorem 8-10. ( P 235 ) Through a given point there passes one and only one lme perpendicular t o a given plane,
Theorem 8-11. ( P . 2 3 5 ) The shortest segment to a plane f r o m an external p o i n t I s the perpendicular segment.
Theorem.
(P. 242)
Two parallel lines l i e in exactly
one plane. ,(F. 242 Two lmes l.n a plane are p a r a l l e l if they are both perpendicular to the same l i n e .
m. (P. 244) point not on L. parallel to L.
Wt L be a line, and l e t P be a Then there is at least one line through P,
Theorem.
(P. 246) If two l i n e s are cut by a transversal,
and if one pair of alternate interior angles are congruent, then the other pair of alternate Interior angles are also congruent. Theorem 9-5. (P. 246) If t w o lines are cut by a transversal, and if a pair of alternate interior angles are congruent, then the lines are parallel. Theorem 9-6. (P. 252) If two lines are c u t by a transversal, and if one pair of corresponding angles are congruent, then the other three p a i r s of corresponding angles have the same property. Theorem 9-7. (P. 252) If two lines are cut by a transversal, and if a pair of corresponding angles are congruent, then the lines are parallel.
Theorem 9-8. (P. 253 ) ~f two parallel l i n e s are cut by a transversal, then then a l t e r n a t e Interior angles are congruent.
Theorem 9-9. (P. 253 ) If two parallel lines are c u t by a transversal, each p a i r of corresponding angles are congruent. Theorem 9-10. ( P . 254) If two parallel lines are c u t by a transversal, I n t e r i o r angles on the same s i d e of t h e transversal are supplementary.
Theorem 9-11. ( P . 255) In a plane, two lines parallel to the same line are parallel to each other. Theorem 9-12. ( P . 2 5 5 ) In a plane, If a l i n e is perpendlcular to one of two parallel lines It l a perpendicular t o the other. Theorem 9-13. (P. 238) The sum of the measures of t h e angles of .a t r i a n g l e is 180.
Corollary 9-13-1. (P. 2 5 9 ) Given a correspondence between two triangles If two pairs of corresponding angles are congruent, then t h e t h i r d pair of corresponding angles are also congruent.
.
Corollary 9-13-2. (P. 260) The acute angles of a right triangle are complementary. Corollary 9-13-3. (P. 260 ) For any triangle, the measure of an exterior angle is the sum of the measures of the two remote interior angles.
.
Theorem 9-14. (P 265 } E i t h e r diagonal d i v i d e s a parallelo- ; gram i n t o two congruent triangles.
Theorem 9-15. ( P . 265 ) sides are congruent.
In a parallelogram, any
two opposite
: 1
Corollary 9-15-1. ( p . 2 6 6 ) If L, 1 1 L, and if P and Q are any two points on L,, then the distances of P and Q from Ly are equal. Theorem 9-16.
(P. 2 6 6 )
in a parallelogram, any two opposite
angles are congruent. Theorem 9-17. ( P . 266) In a parallelogram, any two consecutive angles are supplementary.
Theorem 9-18. ( P . 266) bisect each other.
The diagonals of a parallelogram
Theorem 9-19. ( P . 266) Given a quadrilateral In which both pairs of opposite ides are congruent. Then the quadrilateral is a parallelogram.
Theorem 5-20. (P. 266) If two sides of a quadrilateral are parallel and congruent, then the quadrilateral is a parallelogram. Theorem 3-21. (P. 266) If the diagonals of a quadrilateral bisect each o t h e r , then the quadrilateral is a parallelogram.
Theorem 9-22. (P. 267 ) The segment between the mid-points of two sides of a triangle is parallel to the t h i r d s i d e and half as l o n g as the t h i r d s i d e , Theorem 9-23. (P. 268) If a parallelogram has one right angle, then It has four right angles, and the parallelogram I s a rectangle.
Theorem 9-24. (P. 268 ) In a rhombus, the diagonals are perpendicular to one another.
Theorem 9-25., ( P . 268 ) If the diagonals of a quadrilateral bisect each o t h e r and are perpendicular, then the quadrilateral is a rhombus.
w.(P. 2 7 6 )
If three p a r a l l e l lines i n t e r c e p t congruent segments on one transversal, then they Intercept congruent segments on any other transversal. Corollary 9-26-1. (P. 277 ) If three or more parallel lines intercept c o n w e n t segnents on one transversal, then they i n t e r c e p t congruent segments on any o t h e r transversal.
m. (P. 279 )
The medians of a triangle are concurrent In a p o i n t two-thirds the way from any v e r t e x t o the mid-point of the opposite side.
,
Theorem 10-1. (P. 292) If a plane i n t e r s e c t s two parallel planes, then it.intersects them In two parallel lines.
Theorem 10-2. (P. 292) If a line is perpendicular to one of two parallel planes it is perpendicular t o the o t h e r .
Theorem 10-3. (P. 293) same line are parallel.
Two p l a n e s perpendicular t o the
Corollary 10-3-1. (P. 294) If t w o planes are each parallel t o a third plane, they are parallel to each other. Theorem 10-4. (P. 294) plane a r e parallel.
Two lines perpendicular to the same
Corollary 10-4-1. ( I ? . 2 9 4 ) A plane perpendicular to one of two p a r a l l e l l i n e s i s perpendicular to the o t h e r ,
Corollary 10-4-2. (P. 294) If two lines are each parallel to a third they are parallel to each other.
Theorem 10-5. equidistant
.
(P. 295 )
Two parallel planes are everywhere
Theorem 10-6. (p.501) Any two plane angles of a given dihedral angle are congruent.
if a l i n e is perpendicular to a plane, then any plane containing t h i s line is perpendicular to the given plane. Corollary 10-6-1.
(P. 302 )
Corollary 10-6-2. ( p . 3 0 2 ) if two planes are perpendicular, then any line in one of them perpendicular to t h e i r line of intersection is perpendicular to the other plane. Theorem 10-7. (P. 307 ) The projection o f a line i n t o a plane is a l i n e , unless the line and t h e plane are perpendicular. Theorem 11-1. (P. 328 ) the product of Its legs
.
The area of a right t r i a n g l e is h a l f
Theorem 11-2. (P. 328) The area of a t r i a n g l e i s half t h e product of any base and the altitude to t h a t base. Theorem 11-3. (P. 530) The area of a parallelogram I s the product of any base and the corresponding altitude. Theorem 11-4. (P. 333) The area of a trapezoid is half t h e product of i t s altitude and the sum of its bases.
Theorem 11-5. (P. 332) If two triangles have the same altitude, then the ratio of their areas is equal to the r a t i o of their bases.
Theorem 11-6. (P. 3 3 2 ) If two triangles have equal altitudes and equal bases, then they have equal areas. ,Theorem 11-7. (P. 339) h he Pythagorean Theorem.) In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs.
Theorem 11-8. (P. 340) If the square of one s i d e of a triangle I s equal to the sum of the squares of the other two, then the triangle is a right triangle, with a right angle opposite the f i r s t s i d e . Theorem 11-9. (P. 346) (The 30 - 60 Triangle Theorem.) The hypotenuse of a right triangle I s twice as long a s the shorter l e g if and only if the acute angles are 30' and 60". Theorem 11-10. ( P . 346 ) (The Isosceles R i g h t Triangle he or em.) A right triangle I s isosceles if and only If t h e hypotenuse I s -/;T times as long as a l e g .
.
Theorem 12-1. (P 368 ) he Basic Proportionality heo or em .) If a line parallel to one s i d e of a triangle Intersects the other two sides in d i s t i n c t points, then I t c u t s off segments which are proportional to these s i d e s .
Theorem 12-2. (P. 369 ) If a line intersects two sides of a triangle, and c u t s o f f segments proportional to these two s i d e s , then it I s parallel to the third side. Ã
Theorem 12-3. ( p . 3 T 4 ) h he A . A . A . Similarity he or em.) Given a correspondence between two triangles. If corresponding angles are congruent, then the correspondence is a similarity. Corollary 12-3-1. (P,376 ) h he A .A. Corollary.) Given a correspondence between two triangles. If two pairs of corresponding angles are congruent, then the correspondence is a similarity.
Corollary 12-3-2. (P. 376) If a line parallel to one side of a triangle intersects the other two sides in distinct points, then it cuts o f f a triangle similar t o t h e given triangle.
.
Theorem 12-4. (P, 376) (The S .A .S Similarity heo or ern.) Given a correspondence between two triangles. If two corresponding angles are congruent, and the including sides are proportional, then the correspondence is a similarity.
Theorem 12-5. (P. 3788) h he S .S.S. Similarity Theorem.) Given a correspondence between two triangles. If corresponding sides are proportional, then t h e correspondence i s a similarity. Theorem 12-6. (P. 391) In any right t r i a n g l e , the altitude to the hypotenuse separates the triangle i n t o two triangles which are similar both to each other and to the o r i g i n a l triangle. Corollary 12-6-1. (P. 392) Given a r i g h t triangle and the altitude from the r i g h t angle to the hypotenuse: (1) The altitude is the geometric mean of the segments i n t o which it separates t h e hypotenuse.
(2) E i t h e r leg Is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to the leg. Theorem 12-7. f P . 5 ~ 5 ) The ratio of the areas of two similar triangles I s the square of t h e r a t i o of any two corres-
ponding sides.
.
(P 410) The intersection of a sphere w i t h a p l a n e t h r o u g h I t s c e n t e r is a c i r c l e w i t h the same c e n t e r and Theorem 15-1.
radius.
.
Theorem 13-2. ( P 414) Given a l i n e and a c i r c l e in the same p l a n e . L e t P be t h e center of the c i r c l e , and l e t F be the foot of t h e perpendicular from P to t h e lineà Then e i t h e r ( 1 ) Every p o i n t of the l i n e is outside the circle, or (2) F is on t h e c i r c l e , and t h e l i n e is t a n g e n t to t h e circle a t F, or ( 3 ) F is inside t h e c i r c l e , and the l i n e intersects t h e c i r c l e in exactly t w o p o i n t s , which a r e equidistant from F.
.
( ? 416) Every line tangent to C is p e r p e n d i c u l a r to the radius drawn to t h e p o i n t of c o n t a c t .
Corollary 13-2-1.
Corollary 13-2-2. (P. 416) Any line in E, perpendicular to a radius a t its outer end, is t a n g e n t t o t h e c i r c l e .
Corollary 13-2-3. ( P . 416) Any perpendicular from t h e c e n t e r of C t o a chord bisects the chord. Corollary 15-2-4. ( P . 416) The segment joining the center of C to the mid-point of a chord is perpendicular to the chord.
I n t h e plane of a circle, the p e r p e n d i c u l a r b i s e c t o r of a chord passes through the center of the circle. Corollary 13-2-5.
( P . 416)
Corollary 13-2-6. (P. 417) If a line in the plane of a circle Intersects the interior of t h e c i r c l e , then i t I n t e r s e c t s t h e c i r c l e in
exactly two p o i n t s ,
Theorem 13-3. ( P . 417) In the same circle or in congruent circles, chords equidistant from t h e center are congruent. Theorem 13-4. (P. '417) In the same c i r c l e or in congruent c i r c l e s , any two congruent c h o r d s are e q u i d i s t a n t from t h e c e n t e r .
Theorem 13-5. (F. 424) Given a p l a n e E and a sphere S with c e n t e r P. Let P be t h e f o o t of the p e r p e n d i c u l a r segment from P to E. Then either (1) Every p o i n t of E is o u t s i d e S , or ( 2 ) F is on S, and E is tangent t o 3 at F, or ( 3 ) F is I n s i d e S, and E i n t e r s e c t s S in a c i r c l e w i t h center
P.
C o r o l l a r y 13-5-1. (P. 426) A plane tangent to S I s perpendicular to the radius drawn to t h e point of contact.
Corollary 13-5-2. (P. 426) A plane p e r p e n d i c u l a r t o a r a d i u s at i t s o u t e r end is tangent to S .
Corollary 13-5-3. (P. 426) A perpendicular from P to a chord of S b i s e c t s the chord.
Corollary 15-5-4. (P. 426)
The segment joining t h e c e n t e r of S to t h e midpoint of a chord is perpendicular to the chord.
a
.
Theorem 13-6. (P 431 ) If and @ are arcs of the same circle having only t h e point B In common, and if their union is an a r c "St?, then m'ftS + r&! =
ma.
Theorem 13-7. ( P . 434) The measure of an inscribed angle I s h a l f the measure of its i n t e r c e p t e d a r c .
Corollary 13-7-1 . (P c i r c l e is a right angle.
. 437 )
Corollary 13-7-2. (P. 437) arc are congruent.
An angle inscribed In a semi-
Angles inscribed in the same
Theorem 13-8. (P. 441) In the same c i r c l e or in congruent circles, If two chords are congruent, then so a l s o a r e the
corresponding minor a r c s . Theorem 13-9. ( P . w1) In t h e same circle o r i n congruent c i r c l e s , If two arcs are congruent, then so are the c o r r e s p o n d i n g chords
.
-Theorem 15-10. ( P . 442) Given an angle with v e r t e x on the c i r c l e formed by a secant ray and a tangent ray. The measure of the angle I s h a l f the measure of the Intercepted a r c . Theorem 13-11. (P. 448) The t w o tangent segments to a c i r c l e f r o m an external point are congruent, and form congruent angles with the l i n e Joining the external point t o the c e n t e r of the c i r c l e .
.
Theorem 15-12. ( P 449) G i v e n a circle C and an external point Q, let L1 be a secant line through Q, intersecting C i n p o i n t s R and S ; and l e t L p b e another secant line through Q, intersecting C in p o i n t s T and U. Then QR QS = QU QT.
Theorem 13-13. (P. 450) Given a tangent segment to a c i r c l e , and a secant l i n e through Q, intersecting the c i r c l e in points R and 3 . Then QR QS = Q T .
Theorem 13-14,
(P. 451) If two c h o r d s i n t e r s e c t w i t h i n a
circle, t h e product of t h e lengths of t h e segments of one equals t h e product o f t h e lengths of t h e segments of t h e other.
Theorem 14-1. (P . 467) The bisector of an angle, minus its end-point, is t h e s e t of points in t h e interior of the angle equidistant from t h e sides of t h e angle.
.
Theorem 14-2. (P 469) The perpendicular "bisectors of t h e s i d e s of a t r i a n g l e are concurrent in a point e q u i d i s t a n t from t h e three vertices of t h e t r i a n g l e .
Corollary 14-2-1. (P. 470) There is one and only one c i r c l e through t h r e e non-coll lnear polnfcs
.
Corollary 14-2-2. (P. 470) i n t e r s e c t in a t most two points.
Theorem 14-3.
(P. 470)
Two d i s t i n c t c i r c l e s can
The three altitudes of a triangle
are c o n c u r r e n t . Theorem 14-4. ( P . 471) The a n g l e bisectors of a t r i a n g l e are c o n c u r r e n t in a point e q u i d i s t a n t from t h e t h r e e sides. Theorem 14-5. (P. 476) (The Two C i r c l e he or em) If two circles have radii a and b, and if c is the distance between t h e i r centers, then the c i r c l e s intersect in two p o i n t s , one on each side of the l i n e of centers, provided each of of a, b , c I s l e s s than t h e sum of t h e other two. Construction 14-6.
(P. 477)
TO
C o n s t r u c t i o n 14-7.
( P . 479)
To copy a given angle.
Construe tJ.on 14-8. (P . 481) b i s e c t o r of a given segment.
Corollary 14-8-1.
Construction 14-9.
.
(P 481 )
.
copy a given triangle.
To construct the p e r p e n d i c u l a r To b i s e c t a given segment
.
(P 482)
To c o n s t r u c t a perpendicular to a given line through a given point. Construction 14-10. (P. 484) To construct a p a r a l l e l t o a given line, through a given external point.
Construction 14-11.
.
To d i v i d e a segment into a
(F 484)
given number of congruent segments.
C o n s t r u c t i o n 14-12. a given t r i a n g l e . C o n s t r u c t i o n 14-15.
Construction 1k-14. given t r i a n g l e .
( P . 491)
To circumscribe a c i r c l e about
(P . 491 )
To b i s e c t a g i v e n a n g l e .
(P. 492)
To inscribe a c i r c l e in a
Theorem 15-1. (P. 517) The ratio F ,of the circumference to t h e diameter, Is tlie same for a l l c i r c l e s . Theorem 15-2.
i s Trr
2
(P. 522)
The area of a c i r c l e of radius r
Theorem 15-3. (P. 526) If two arcs have equal radii, their l e n g t h s a r e proportional t o t h e i r measures. Theorem 15-4. has length &qr
(P
.
. 526)
An
arc of measure q and radius r
Theorem 15-5. (P. 527) The area of a s e c t o r is h a l f t h e product of i t s radius by the l e n g t h of i t s a r c . Theorem 15-6. (P. 527) The area of a sector of radius 2 and a r c measure q is ^yip
r
.
Theorem 16-1. (P. 535) All cross-sections of a t r i a n g u l a r p r i s m are congruent to t h e base.
Corollary 16-1-1. (P. 536) The upper and lower bases of a t r i a n g u l a r prism are congruent. Theorem 16-2. ( P
. 536)
.
( p r i s m Cross-Section heo or em )
All
cross-sections of a prism have t h e same area.
.
Corollary 16-2-1. ( P 537)
The two bases of a prism have
equal areas. Theorem 16-3. (P. 537) The l a t e r a l f a c e s o f a prism are parallelogram regions, and t h e l a t e r a l faces of a right prism are r e c t a n g u l a r regions.
16-4.
.
(P 540) -A cross-section of a triangular pyramid, by a plane between the vertex and the base, Is a t r i a n g u l a r region similar to the base. If the distance from the v e r t e x t o the cross-section plane is k and the altitude is h, then the r a t i o of t h e area of t h e cross-section t o the area of k the base I s Theorem
.
Theorem
16-5, (P. 542) In m y pyramid, t h e ratio of the
of a cross-section and t h e area of the base Is ($) , where h i s t h e a l t i t u d e of the pyramid and k is the d i s t a n c e from the vertex to the plane of the cross-section. a
Theorem
.
16-6, (P 543) (The Pyramid Cross-Section
heo or em. )
Given two pyramids w i t h the same a l t i t u d e . If the bases have the same area, then cross-sections equidistant from the bases a l s o have t h e same area.
.
Theorem 16-7. (P 548) The volume of any prism is the product of the altitude and the area of the base. Theorem 16-8. (P. 549) If two pyramids have t h e same altltude and the same base area, then they have the same volume.
Theorem 16-9. (P. 550) The volume of a t r i a n g u l a r pyramid is one-third t h e product of its a l t i t u d e and i t s base area.
.
Theorem 16-10. (P 551) The volume of a pyramid is o n e - t h i r d t h e product of its a l t i t u d e and i t s base a r e a .
Theorem 16-11. ( P . 555) A cross-section of a c i r c u l a r cylinder is a c i r c u l a r r e g i o n congruent t o t h e base.
Theoren 16-12. ( P . 555) The area of a cross-section of a c i r c u l a r c y l i n d e r is equal to t h e area of the base, Theorem 16-13. (P. 555) A cross-section of a cone of a l t i t u d e h, made by a plane at a distance k from t h e vertex, is a c i r c u l a $ region whose area has a r a t i o to the area of the k base of (,-)
.
.
Theorem 16-14. (P 557) The volume of a c i r c u l a r cylinder is the product of the altitude and the area o f the base.
.
Theorem 16-15. (P 557) The volume o f a c i r c u l a r cone Is one-third t h e product of t h e a l t i t u d e and t h e area of t h e base.
Theorem 16-16. (P. 559) The volume of a sphere of radius
P
+2.
is
Theorem 16-17. radius r is S =
(P. 562) The surface area of a sphere of
.
h2
Theorem 17-1. (P. 579) have the same s l o p e .
On a non-vertical l i n e , all segments
Theorem 17-2. (P. 584) Two non-vertical lines are p a r a l l e l if and only I f they have the same slope.
.
Theorem 17-3. (P 586) Two non-vertical l i n e s a r e perpend i c u l a r if and o n l y if their slopes are the negative reciprocals of each o t h e r . Theorem 17-4. (P. 589) (The Distance p or mu la.) The distance between the p o i n t s (x,,y-,) and (xpy2) is equal t o
Theorem 17-5. (P. 593) h he Mid-Point Fo??nIulaO) ~ e tP^ = ( x f i ) and l e t Pg = ( x g 9 y 2) . Then the mid-point
-
of PIPn
i s the point
x + P = ( 1 2
.
X^
Y ! 1
+
2
y2
1
Theorem 17-6. ( P 605) L e t L be a non-vertlcal l i n e w i t h w i t h slope m, and let P be a point of L, w i t h coordinates (xl,yl), For every p o i n t Q = ( x , ~ ) of L, t h e e q u a t i o n is s a t i s f i e d . y yi = m(x - x,)
-
Theorem 17-7. ( P . 607) The graph of the equation is the line that passes through the p o i n t y y- = m ( x (x,,y.,) and has slope m.
-
x.,)
Theorem 17-8. ( P . 611) The g r a p h of the equation y = mx is t h e line w i t h slope m and y - i n t e r c e p t b.
+
b
Theorem 17-9. (P. 613) Every l i n e I n the plane is t h e graph of a l i n e a r equation in x and y.
Theorem 17-10. ( P . 615) The graph of a linear e q u a t i o n in x and y is always a l i n e . Theorem 17-U. (x
-
a)2
+
and r a d i u s
(y
-
(P. 6 2 3 ) The graph of t h e e q u a t i o n
b)* = r2
Is the c i r c l e w i t h c e n t e r at
(a,b)
r.
Theorem 17-12. equation of t h e form
(P. 624) Every c i r c l e I s t h e graph of an s2
+
y2
Theorem 17-13. (P. 625)
+
Ax
+
By
+
C
=
0.
Given t h e equation
x2 + y2 + Ax + By + C = 0. The graph of this equation Is (1) a c i r c l e , (2) a p o i n t or ( 3 ) the empty s e t .
Index of --
Definitions
For precisely defined geometric terms the reference I s to the formal definition. For other terms the reference is to an informal definition or t o t h e most prominent d i s c u s s i o n .
absolute value, 27 acute angles, 86 a l t e r n a t e i n t e r i o r angles, 245 altitude
of prism, 535 of pyramid, 540 of triangle, 214, 215 angleb), 71 acute, 86 alternate Interior, 245 bisector o f , 129 central, 429 complementary, 86 congruent, 86, 109 consecutive, 264 corresponding, 251 d i h e d r a l , 299 exterior, 293 exterior of, 75 i n s c r i b e d , 432 intercepts an arc, 433 i n t e r i o r o f , 73 measure of, 79, 80 obtuse, 86 or polygon, 506 opposite, 264 reflex, 78 remote i n t e r i o r , 195 r i g h t , 85 right d i h e d r a l , 301 sides of, 71 straight, 78 supplementary, 82 vertex of, 71 vertical, 88 apothem, 512 a r c ( s ) , 429 center of, 437 congruent , 441 degree measure of, 430 end-points of, 429 length of, 525 major, 429 minor, 429
of sector, 527
area, 320
,,
circle, 521, 522 parallelogram, 330 polygonal region, 320 rectangle, 322 right t r i a n g l e , 328 sphere, 562 trapezoid, 351 triangle, 328 unit of, 321 arithmetic mean, 364 auxiliary s e t s , 176 base of pyramid, 540 between, 41, 182 bisector of an angle, 129 bisector of a segment, 169 b i s e c t s , 47, 129 Cavalleri~sPrinciple, 548
center of arc, 437 c i r c l e , 409 sphere, 409
central angle, 429
centroid, 280, 621 chord. 410 c i r c l e ( s ) , 409 a r e a of, 521, 522 circumference of, 516 congruent, 417 equation o f , 623, 624, 625 exterior of, 412 great, 410 Interior of, 412 segment of, 528 tangent, 417
circular cone, 554
cylinder, 553 reasoning, 119 region, 520 area of, 521 circumference , 516
circumscribed
circle, 490 triangle, 490 colllnear, 54 complement, 86 complementary angles, 86 concentric circles, 409 spheres, 409 conclusion, 60
concurrent sets, 278, 469 cone,
c i r c u l a r , 554
r i g h t circular,
volume of, 557
555
congruence , 97
congruent
angles, 86, 109
arcs, 441 c i r c l e s , 43.7 segments, 109 triangles, 98, 111 consecutive angles, 264 consecutive s i d e s , 264 cons t r u c t ions, 477 converse, 202 convex polygon, $07 convex sets, 62 coordinate system, 37, 571 coordinates of a point, 37, 569 co-planar, 6 4 corollary, 128 correspondence, 97 corresponding angles, 251 cross-section of a prism, 535 of a pyramid, 540 cube, 229 cylinder
circular, 553
volume o f , 557 diagonal, 264, 509 diameter, 410 d i h e d r a l angle, 299 edge of, 299 f a c e of, 299 measure of, 301 p l a n e angle o f , 300
distance, 34 distance between a point and a l i n e , 206 a point and a plane, 235 two parallel l i n e s , 266 distance formula, 589 edge of half plane, 64 end-point ( s ) of a r c , 429 of r a y , 46 of segment, 45 empty s e t , 18
equation of c i r c l e , 623 of l i n e , 605, 611 e q u i a n g u l a r triangle, 128 e q u i l a t e r a l t r i a n g l e , 128 Euler, 327 existence p r o o f s , 165 e x t e r i o r angle, 193 ex'ce-'-Lor
o r an
a n g l e , 73 o f a c i r c l e , 412 of a t r i a n g l e , 7 G f ~ c eof h a l f - s p a c e , 66
frustum, 559
Garfield's Proof, 344 geometric mean, 561 graph, 600 great c i r c l e , 410 half-plane, 6Q edge of 64 ha1 f -space, 66 f a c e o f , 66 h o r i z o n t a l l i n e s , 576 hypotenuse, 172 h y p o t h e s i s , 60 identity congruence, 100, 109 i f and only if, 203 i f - t h e n , 60 inconsistent e q u a t i o n s , 618 i n d i r e c t proof, 160 Inequalities, 24 i n f i n i t e r u l e r , 37 inscribed
angle, 432
measure o f , 434
c i r c l e , 490 polygon, ,511 quadrilateral, 438 t r i a n g l e , 490 i n t e g e r s , 22 i n t e r c e p t , 275, 433
interior of a n g l e , 73
of circle, 412 of triangle, 7 4 i n t e r s e c t , 18 intersection of sets, 16, 18, 4-73 i r r a t i o n a l numbers, 23 isosceles triangle, 127 kite, 272
lateral edge, face,
53'7
537
s u r f a c e , 537 lemma, 196 length of a r c , 525 of segment, 45 linear equation, 613 linear pair, 82 line(s), 10 oblique, 216 p a r a l l e l , 241
perpendicular, 86 skew, 241
transversal, 244 major a r c , 429 mean arithmetic, 364 geometric, 361 measure of angle, 7 9 , 80
of dihedral angle, 301 of distance, 5 0 , 54, 36 median of t r a p e z o i d , 272 o f triangle, 130
mid-point, 47 f o r m u l a of, 595 minor arc, 429 Non-Euclidean geometries, 255 n e g a t i v e real numbers, 191 numbers
Irrational, 23 negat i v e , 191 p o s i t i v e , 191 rational, 22
r e a l , 25 whole, 22 o b l i q u e l i n e s , 216 obtuse angle, 86 on o p p o s i t e sides, 64
on the same s i d e , 64 one-to-one correspondence, 97 opposite angles 264 rays, 46 sides, 264 order, 24 order p o s t u l a t e s , 191, 192
ordered pair, 571 origin, 568 parallel l i n e s , 241 slopes of, 584 l i n e s and p l a n e s , 291 planes, 291 parallelepiped, 538 parallelogram, 265 area of, 330 perimeter of triangle, 287 of polygon, 512" perpendicular l i n e s . 86 slopes of, 586 line and plane, 219 planes, 301 p e r p e n d i c u l a r b i s e c t o r , 169 PI,
TT,
518
plane(s), 10 p a r a l l el, 291 perpendicular, 301 plane angle, 300 point, 10 p o i n t - s l o p e form, 605 p o i n t of ';angency of c i r c l e s , 413 of spheres, 423 polygon, 506 a n g l e o f , 506 apothem of, 512
convex, 507 diagonal of, 509
inscribed, 511 perimeter of, 512 regular, 511 sides o f , 506 v e r t i c e s of, 506 polygonal r e g i o n , 317 p o l y h e d r a l regions, 546 positive real numbers, 191 postulate ( s ), 9 of order, 191, 192 power* of a p o i n t , 451 prism, 554 a l t i t u d e o f , 535 cross-section of, 535 l a t e r a l edge, 537 l a t e r a l f a c e , 537 1a t e pal surface , 537 lower base, 555 r e c t a n g u l a r . 555
prism (continued)
535 t o t a l surface, 537 t r i a n g u l a r , 535 upper base, 535 projection of a l i n e , 306 right,
proof
of a point;, 506
converse, 202 double-column form of, 118 existence, 165 i n d i r e c t , 160 uniqueness , 165 writing of, 117
p r o p o r t i o n a l sequences, 360
pyramid, 540 altitude of, 540 base o f , 540 regular, 5 4 4 vertex of, 540 volume o f , 551
Pythagorean Theorem, 339
q u a d r a n t , 571
q u a d r i l a t e r a l , 263 consecutive angles of, 264 consecutive s i d e s of, 264 cyc l LC,
473
diagonal of, 264 i n s c r i b e d , 438 opposite angles of, 264 radius, 409, 410 of sector, 527 r a t i o n a l numbers, 22 ray, 46 end-point of, 46 opposite, 46 real numbers, 25 rectangle, 268 area o f , 322 rectangular p a r a l l e l e p i p e d , 538 reflex angle, 78 region c i r c u l a r , 520 polygonal, 517 polyhedral, 546 t r i a n g u l a r , 317 regular polygon, 511 pyramid, 544 remote i n t e r i o r angle, 193
rhombus, 268 r i g h t angle, 85 , right d i h e d r a l angle, 301 right prism, 535 right t r i a n g l e , 172 s c a l e n e triangle, 128 s e c t o r , 527 arc of, 527 r a d i u s of, 527 segment ( s ), 45 b i s e c t o r , 169 congruent , 109 segment of a c i r c l e , 528 semi-circle, 429 separation, 182 sefc(s), 15 auxiliary, 176 concurrent, 278 convex, 62 element o f , 15 empty, 18 intersection of, 16, 473 member of, 15 union, of, 17 side ( a ) c o n s e c u t i v e , 364 of angle, 71 of d i h e d r a l angle, 299 of polygon, 506 of t r i a n g l e , 72 opposite, 264 similarity, 365 skew l i n e s , 241 slope, 577 of parallel lines, 5811 of perpendicular lines, 586 slope-intercept form, 611 space, 53 sphere, 409 e x t e r i o r of, 423 interior of, 425 surface area of, 562 volume of, 559 square, 268 square root, 25 straight angle, 78 s u b s e t , 15 supplement, 82 supplementary angles, 82 -
-
tangent circles, 417
common external, 454 common i n t e r n a l , 454 externally, 417 internally, 417 l i n e and c i r c l e , 413
plane and s here, 423 segment., 44
1
theorem, 9 total surface o f a prism, 537 transversal, 244 trapezoid, 265 area of, 331 t r i a n g l e ( s ) , 72 a l t i t u d e of, 214 a n g l e bisector of, 130 area of, 328 c e n t r o i d of, 280 congruent, 98, 3-11 equiangular, 128 equilateral, 128
e x t e r i o r o f , 74
Interior of, 74 isosceles, 127, 128, 346 median o f , 130 overlapping, 123 perimeter o f , 287 right, 172
scalene, 128 s i d e s of, 72 s i p l a g , 365 30 -60 , 346 vertex of, 72 triangular region, 317 undefined terns, 9 , 10 union o f s e t s , 17 uniqueness proofs, 165 vertex of angle, 71 of polygon, 506 of pyramid, 540 of triangle, vertical angles, v e r t i c a l l i n e , 576
2
volume of cone, 557 of cylinder 557 of prism, 5 k of pyramid, 551 of sphere, 559 whole numbers, 22 x-axis, 568 y-axis, 568 y-intercept, 611
