IMPORTANT THEOREMS AND PROOFS Theorem: If a perpendicular is drawn from the vertex of the right angle of a right angled triangle to the hypotenuse, the triangle on each side of the perpendicular are similar to the whole triangle and to each other. Given: In a triangle ABC, right angled at B, BD ⊥ AC
To prove: Triangle ADB Similar to Triangle BDC
Triangle ADB Similar to Triangle ABC Triangle BDC Similar to Triangle ABC Proof: (1) In Triangle ABC , ∠B = 90o this implies that ∠ABD + ∠BC = 90o In Triangle DBC ,sin ce∠D = 90o
Therefore ∠DBC + ∠BCD = 90o So, ∠ABD = ∠BCD In Triangle ADB and Triangle BDC we have
∠ABD = ∠BCD and ∠ABD = ∠BCD = 90o
(AA-similarity)
Therefore Triangle ADB Similar to Triangle BDC
(2) In Triangle ADB and Triangle ABC , ∠ABC = ∠ADB = 90o and ∠A is common
Therefore Triangle ADB Similar to Triangle ABC (AA-similarity) (3) In Triangle BDC and Triangle ABC , ∠BDC = ∠ABC = 90o and ∠C is common
Therefore Triangle BDC Similar to Triangle ABC (AA-similarity) Theorem: The ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides. Given: Two triangles, Triangle ABC Triangle ABC Similar to Triangle DEF
and Triangle DEF ,
such
that
AB BC AC = = , ∠A = ∠D, ∠B = ∠E , ∠C = ∠F DE EF DF To prove:
Area of Triangle ABC AB 2 BC 2 AC 2 = = = Area of Triangle DEF DE 2 EF 2 DF 2
Construction: Through A draw AP ⊥ BC and through D draw DQ ⊥ EF
1 1 Proof: Area of Triangle ABC = × BC × AP (area of a Triangle = × base × height 2 2
1 Area Triangle DEF = × EF × DQ 2 1 × BC × AP Area of Triangle ABC 2 BC × AP = = Now 1 Area of Triangle DEF × EF × DQ EF × DQ 2 Area of Triangle ABC ⎛ BC ⎞ ⎛ AP ⎞ =⎜ ⎟ ⎟×⎜ Area of Triangle DEF ⎝ EF ⎠ ⎝ DQ ⎠
… (1)
In Triangle ABP and Triangle DEQ
∠1 = ∠2 = 90o
(Construction)
∠3 = ∠4
(Given)
Therefore Triangle APB Similar to Triangle DQE or
(AA corollary)
AP PB AB = = DQ QE DE AB AP = DE DQ
⇒
or AB BC = DE EF
BC AP (Given) = EF DQ
… (2)
Using (1) and (2), we get
Area of Triangle ABC ⎛ BC ⎞ ⎛ AP ⎞ =⎜ ⎟ ⎟×⎜ Area of Triangle DEF ⎝ EF ⎠ ⎝ DQ ⎠ Area of Triangle ABC ⎛ BC ⎞ ⎛ BC ⎞ BC 2 =⎜ or ⎟×⎜ ⎟= Area of Triangle DEF ⎝ EF ⎠ ⎝ EF ⎠ EF 2 In the same manner, it can be proved that
… (3)
Area of Triangle ABC AB 2 = Area of Triangle DEF DE 2
… (4)
Area of Triangle ABC AC 2 = Area of Triangle DEF DE 2
… (5)
Combining (3), (4) and (5), we get
Area of Triangle ABC AB 2 AC 2 BC 2 = = = Area of Triangle DEF DE 2 DE 2 EF 2 Theorem: (PYTHOGRAS THEOREM): In a right-angled triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides. Given: Triangle ABC is right angled at A, that is ∠BAC = 90o To prove: BC 2 = CA2 + AB 2 Construction: Through A draw AD ⊥ BC Proof: In Triangle ABC and Triangle DBA , we have
∠ABC = ∠ABD
(Common to both the triangles)
∠BAC = ∠ADB = 90o Therefore Triangle ABC Similar to Triangle DBA (AA-criterion of similarity) We know that in similar triangles, corresponding sides are proportional. Therefore
or
AB BC AC = = DB BA DA
BC AB = BA DB
Cross-multiplying both sides, we get AB × BA = BC × DB
or AB 2 = BC × DB
… (1)
Now in Triangle ABC and Triangle ADC
∠CAB = ∠ADC = 90o ∠C = ∠C
(Common to both the triangles)
Therefore By AA criterion of similarity, we have Triangle ABC Similar to Triangle ADC Therefore
Taking
BC AB AC = = AC AD DC
AC BC and cross multiplying, we have = DC AC
AC 2 = BC × DC Adding (1) and (2), we get AB 2 + AC 2 = ( BC × DB) + ( BC × DC ) AB 2 + AC 2 = BC ( DB + DC )
… (2)
AB 2 + AC 2 = BC × BC
[ Since DB + DC = BC ]
AB 2 + AC 2 = BC 2 Theorem: (Converse of Pythagoras theorem): In triangle, if the square o one side is equal to the sum of the squares on the remaining two sides, then the angle opposite t the first side is a right angle. Given: Triangle ABC , such that AB 2 + AC 2 = BC 2 To prove: ∠ACB = 90o that is Triangle ABC , is a right triangle. Construction: Draw a right triangle DEF such that BC=EF, DF=AC and ∠EFD = 90o
Proof: Since ∠EFD = 90o therefore Triangle DEF is right angled triangle. By Pythagoras theorem, we have
DE 2 = EF 2 + FD 2 EF=BC FD=AC
Therefore DE 2 = BC 2 + AC 2
… (1)
Also AB 2 = BC 2 + AC 2
… (2)
So, DE 2 = AB 2
DE = AB Now in Triangle ABC and Triangle DEF , we have AB=DE (proved above) BC=EF (given) AC=DF (given)
Therefore Triangle ABC ≅ Triangle DEF
(SSS-criterion of congruence)
As corresponding parts of congruent triangles are congruent,
Therefore ∠ACB = ∠EFD But ∠EFD = 90o
(By construction)
Therefore ∠ACB = 90o Hence Triangle ABC is right angled triangle, right angled at C. Theorem: The internal bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle. Given: Triangle ABC in which AD is the internal bisector of ∠A and meets BC in D. To prove:
AB BD = AC DC
Construction: Draw CE parallel to DA , meeting BA produced in E.
Proof: Since AD parallel to CE
Therefore ∠2 = ∠3 ∠1 = ∠4
(Alternate angles) and (Corresponding angles)
But ∠1 = ∠2 Since AD bisects ∠A
Therefore ∠3 = ∠4 So AE=AC Now, in Triangle BCE , DA parallel to CE
Therefore
Hence
BA BD = AE DC
AB BD = AC DC
(By Basic Proportionality Theorem)
( Since AE = AC )
Theorem (Converse of Previous Theorem): If a line through one vertex of a triangle divides the opposite sides, in the ratio of other two sides, then the line bisects the angle at the vertex.
Given: Triangle ABC , in which D is a point on BC such that D divides BC in the ratio AB:AC i.e
BD AB = DC AC
To prove: AD is the bisector of ∠A . Construction: Produce BA to E such that AE=AC Proof: In Triangle ACE , we have AE=AC
(by construction)
∠3 = ∠4
… (1)
It is given that
BD AB = DC AC So,
BD AB = DC AE
( Since AC = AE )
Therefore In Triangle BCE , we have BD BA = DC AE
Therefore By converse of basic proportionality theorem we have DA parallel to CE
∠1 = ∠4
(Corresponding angles)
… (2)
and ∠2 = ∠3
(alternate angles)
… (3)
But ∠3 = ∠4
from (1)
Therefore ∠1 = ∠2
From (2) and (3)
Hence AD is the bisector of ∠A . **************************************
