SLC _ Compulsory Math _ Geometric Deduction

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SLC Model Question SLC _ Compulsory Math _ Geometric Deduction [Short

Questions]

# Section A

a)

lbPsf] lrqdf lrqdf

ABCD df CD Pp7f

;r xf] / /

s'g}g ljGb' } ljGb' P 5 CD sf] s'

eg]

G5 APD + BCP = APB x'G5

egL k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . .

In the given figure, ABCD is a parallelogram and P is any point on CD. Prove that APD + BCP = APB.

b)

ABCD Pp6f

;dfgfGt/ rt'e{ e '{h xf] . .

s'g}g ljGb' } ljGb' P / AD sf] s' s'g}g ljGb' } ljGb' Q 5 AB sf] s'

eg]

QBC = APD + CPB

x'G5 G5 elg

k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . . ABCD is a parallelogram. P is any point in AB and Q is any point on AD. Prove that. QBC = APD + CPB.

c) AXB =

rt'e{ e 'h {

lbOPsf] lrqdf lrqdf

ABCD Pp6f

;=r= xf] h;df h;df ljs0f{

BD df

kg{ ] s' s'g}g ljGb' } ljGb'

X5

. k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . .

ABCX.

In the given figure, ABCD is a parallelogram is which X is any point on the diagonal BD. Prove that : AXB = quadrilateral ABCX.

d)

;Fu} usf] }sf] lrqdf lrqdf

PQRSPp6f

;dfgfGt/ rt'e{ e '{h xf] . . ljs0f{

s'g}g ljGb' } ljGb' M ;Fu Q / S hf]l8Psf l8Psf PR sf] s'

5g\ . . l;4 ug{ 'xf] xf];\ ;\ . .

If]qkmn qkmn = PSM sf] If] If]qkmn qkmn PQM sf] If]

In the adjoining diagram, PQRS is a parallelogram, Q and S are joined to any point M on the diagonal PR of the parallelogram Prove that are of the PQM = area of PSM e)

rt'e{ e '{h

G5 ABCD df AO = Co eP 2ABD = ABCD x'G5


In quadrilateral ABCD, AO = Co, prove that 2ABD = ABCD.

zLif{ljGb' ljGb' A / BC sf] Pp6f Pp6f f) ABC sf] zLif{

ljGb'

l8Psf] 5 5 D hf]l8Psf]

/

dWoljGb' E 5 AD sf] dWoljGb'

eg]

G5 ABC = 2EBC x'G5

elg k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . .

The vertex A of ABC is joined to a point D on the side BC. The midpoint of AD is E, Prove that that ABC = 2EBC.

# Section B

a)

;Fu] usf] ]sf] lrqdf lrqdf

ABCD Pp6f

If]qkmn qkmn = DNM sf] If] If]qkmn qkmn BNC sf] If]

;dfgfGt/ rt'e{ e 'h { xf] . .

CD df

s'g}g} Pp6f Pp6f ljGb'

N5

.

df AN n] BC nfO{ M ljGb'df

5'g]g u/L ] u/L a9fOPsf] 5 5 eg]

x'G5 G5 elg k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . .

In the adjoining figure, ABCD is a parallelogram, N is any point CD AN and BC are produced to meet the point M. Prove that are of BNC = the area of the DNM.



b)

lbOPsf] lrqdf lrqdf

ABCD Pp6f

;=r= xf] . . olb

G5 DC = CQ eP PQC = BPQ x'G5


In the given figure, ABCD is a parallelogram IF DC = CQ, prove that PQC = BPQ.

c)

;=r=

e'hf hf AD df ABCD sf] e'

s'g}g} ljGb' ljGb'

a9fP/ E ljGb'df df F 5 BF / CD nfO{ a9fP/

ldnfOPsf] 5 5 . k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . .

BCE = ABDE.

ABCD is a parallelogram F is any point on side AD.BF and CD are produced to meet at E. Prove that BCE = ABDE.

e'hf hf BC ;Fu ABC sf] e'

d)

;dfgfGt/ /]vf vf

XY 5

.

BE//AC / CF//AB n] XY nfO{ E / F df

e]6\ 65 \ eg]

G5 ABE = ACF x'G5

elg

k|dflf0ft dflf0ft ug{ 'xf] xf];\ ;\ . . XY is a line parallel to side BC of ABC BE//AC and CF//AB meet XY in E and F respectively. Show that ABE = ACF

dWoljGb' D / BC ABC df AB sf] dWoljGb'

e)

sf] s' s'g}g} ljGb' ljGb'

P5

olb .

G5 CQ//DP eP 2BPQ = ABC x'G5

elg k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . .

In ABC, D is the mid-point of AB and P is any point on BC. IF CQ//DP, prove that: 2BPQ = ABC.

f)

lbOPsf] lqe' lqe'h

o{ dlWosfx¿ dlWosfx¿ ABC df, BE / CD b'o{

ljGb'

O df

k|lt5] lt5]lbt lbt 5g\ . . l;4 ug{ 'xf] xf];\ ; . \ .

If]qkmn qkmn = rt'e{ e 'h { ADOE sf] If] If]qkmn qkmn BOC sf] If]

In the given triangle ABC, two medians BE and CD are intersect at O. Prove that area of BOC = area of quadrilateral ADOE.

g)

ABC sf

e'hfx¿ hfx¿

AB, BC, / CA sf

dWoljGb'x¿ x¿ qmdzM

eg] k| k|dfl0ft dfl0ft P, Q, R 5g\ eg]

ug{ 'xf] xf];\ ;\ . .

P, Q and R are the midpoints of sides AB, BC and CA respectively of ABC, prove that, APR = BPQ = CQR = PQR = ABC.

h)

lbOPsf] lrqdf lrqdf

AP, BP / CP sf

dWoljGb'x¿ x¿ qmdzM

\ eg] ABC = 4LMN x'G5 G5 L, M / N x'g eg]

elg k|dfl0ft ug{ 'xf] xf];\ ;\ . .

In the given figure, L, M and N are the mid-points of AP, BP and CP respectively, prove that ABC = 4LMN.

# Section C

a)

rt'e{ e '{h

5f]Psf] Psf] 5 5 . l;4 ug{ 'xf] xf];\ ;\ rt' rt'e{ e '{h

lzif{ljGb' ljGb' C af6 ABCD sf] lzif{

ljs0f{

DB ;Fu

;dfgfGt/ x'g]g u/L ] u/L lvrLPsf] /] /]vfn] vfn]

nDafOPsf] /] /]vfsf] vfsf] ljGb' ljGb' E df AB nfO{ nDafOPsf]

If]qkmn qkmn = DAE sf] If] If]qkmn qkmn ABCD sf] If]

The line drawn through the vertex. C of the quadrilateral ABCD parallel to the diagonal DB meets AB product at E. Prove that the quad ABCD = DAE in area.

b)

lbOPsf] lrqdf lrqdf

PQRS ;dnDa

rt'e{ e '{h xf], h;df

eg] l;4 l;4 PQ//MN//SR 5g\ eg]

ug{ 'xf] xf];\ ;\ . .



If]qkmn qkmn = QRM sf] If] If]qkmn qkmn PSN sf] If] In the given figure, PQRS is a trapezium in which PQ//MN//SR. Prove that area of PSN = area QRM.

# Section D

rlqmo rt'e{ e '{h ABCD df AP, BP, CR / DR qmdzM A, B, C / D sf cw{sx?åf/f sx?åf/f ag]sf] sf] rt' rt'e{ e '{h QPRS klg rlqmo rt'e{ e '{h x'G5 G5

a)

elg k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . . In a cyclic quadrilateral ABCD AP, BP, DR, CR are the bisectors of A, B, D, C respectively. Prove that PQRS is also a cyclic quadrilateral.

b)

lbOPsf] lrqdf lrqdf j[Qsf] Qsf] s] s]Gb| Gb| ljGb' ljGb' O, Jof; AB / DOAB eP AEC = ODA x'G5 G5 elg l;4

ug{ 'xf] xf];\ ; . \ .

In the adjoining figure O is the center of a circle. AB is a diameter and DOAB. Prove AEC=ODA.

c)

lbOPsf] lrqdf lrqdf AD//BC 5 eg] AYC = BXD x'G5 G5 egL k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . .

Given figure is a circle where AD//BC, Prove that AYC=BXD.

d)

lbOPsf] lrqdf lrqdf AC = BC / ABCD Pp6f rlqmo

rt'e{ e '{h xf] . eg] CD n] BDE nfO{ cfwf cfwf ub{5 egL k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . .

In the given figure, AC=BC and ABCD is a cyclic quadrilateral prove that DC bisects BDE.

e)

b'O{O hLjfx? { hLjfx? AB / CD k/:k/ nDa 5g\ AOD + BOC = 1800 x'G5 G5 elg k|dfl0ft dfl0ft ug{ 'xf] xf];\ ; . \ . 0

Two chords AB and CD are perpendicular to each other. Prove AOD+BOC=180 .

f)

lbOPsf] lrq lrq Pp6f j[Qsf] Qsf] xf] xf] . . h;df PMS = QNR eg] PQ//RS x'G5 G5 egL k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . .

Given figure is a circle and PMS = QNR Prove PQ//RS.

g)

lbOPsf] lrqdf lrqdf s]Gb| Gb| ljGb' ljGb' O ePsf] j[ j[Qsf Qsf b'O{O Aof;x? { Aof;x? AB / CD x'g\ g\ . . olb hLjf CE Aof; AB ;Fu

;dfgfGt/ / BOC clws

sf]0f 0f x'g\ g eg] \ eg] rfk rfk DBE sf] dWo dWo ljGb' B x'G5 G5 egL l;4 ug{ 'xf] xf];\ ;\ . . In the figure, AB and CD are two diameters of a circle with center O. If the chord CE is parallel to AB and BOC is obtuse prove that B is the mid point of arc DBE.



h)

lbOPsf] lrqdf lrqdf ABCD Pp6f rlqmo rt'e{ e 'h { xf] . . olb BC = DE / CA n] BCD ;dlåljefhg ub{5 eg] ∆ACE ;dlåjfx' lqe' lqe'h BCD nfO{ ;dlåljefhg

x'G5 G5 egL k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . . In the given figure, ABCD is a cyclic quadrilateral. If BC=DE and CA bisects BCD, prove that ACE is an isosceles triangle.

lbOPsf] j[ j[Qdf Qdf AB Aof; xf] . . OC / OD Aof;fw{x? x?

i)

x'g\ g\, hxfF rfk rfk BC = rfk CD 5 eg] AD//OC x'G5 G5 egL k|dfl0ft dfl0ft ug{ 'xf] xf];\ ; . \ .

In the adjoining figure, AB is the diameter. OC and OD are radii where Arc BC= Arc CD, prove that AD//OC.

# Section E

a)

lbOPsf] j[ j[Qdf Qdf AB Aof;, OC / OD Aof;fy{x? x? x'g\ g\, hxfF BOC =COD

x'g\ g eg] \ eg] AD//OC x'G5 G5 egL l;4 ug{ 'xf] xf];\ ; . \ .

In the adjoining figure, AB is the diameter. OC and OD are radii where BOC=COD, prove that AD//OC.

lrqdf BC, DBE sf] cw{ cw{s b) lbOPsf] lrqdf

xf] . . ABCD rlqmo rt'e{ e '{h xf] eg] eg] k| k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ AC = DC

In the given figure BC is a bisector of DBE. ABCD is a cyclic quadrilateral Prove that AC=DC.

O / P s]Gb| Gb|ljGb' ljGb' ePsf ePsf

c)

b'O{O j[ { j[Qx? Qx? A / B df k|ltR5] ltR5]lbt lbt ePsf 5g\ . . A / B tyf O / P hf]8\ 8bf b\ f C df sf6LPsf 5g\ k| k|dfl0ft dfl0ft

ug{ 'xf] xf];\ ; . \ . i) AC = BC

0

ii) OCA = 90

Two circles having center O and P intersects at points A and B. AB meets OP at point C. Prove that (I) AC=BC (II) OCA=90

d)

Gb|ljGb' ljGb' ePsf] ePsf] j[ j[Q XPY / MQN nfO{ /] /]vf vf XY O s]Gb|

n] X / Y tyf M / N df sf6]sf] sf] 5 5 k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . .

0.

XM = YN

O is the center of two circles XPY and MQN. XY cuts the circle XPY and MQN at X,Y and M,N respectively. Prove XM=YN.

e)

;Fu} usf] }sf] lrqdf lrqdf ljGb' A af6 AB / AC :kz{

/]vfx? vfx? lvlrPsf 5g\ . . :kz{ /] /]vf vf DE ljGb'

F df

:kz{ ePsf ePsf 5g\ eg] eg]

k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . . In the adjoining figure A is the external point of the circle. AB, AC and DE are tangents of the circle at B, C and F. prove that AB+AC=AD+DE+AE. f)

lrqdf AB, BC, CD / DA :kz{ /]vfx? vfx? j[Qsf] Qsf]

3]/fsf] /fsf] qmdzM qmdzM P,Q,R / S

:kz{ ljGb' ljGb'df df :kz{ ePsf

5g\ k| k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ AB + CD =

BC + AD In the given figure AB, BC, CD and DA are tangents of the circle at P, Q, R, S respectively. Prove that AB+CD=BC+AD.



g)

lbOPsf] lrqdf lrqdf b'O{O j[ { j[Qx? Qx? M / N df k|ltR5] ltR5]lbt lbt 5g\ . . PQ / N df ljGb' M af6 kf/ ePsf 5g\ . . R,P,S / Q qmdzM ljGb' N

;Fu hf]l8Psf l8Psf 5g\ eg] eg] l;4 ug{ 'xf] xf];\ ;\ . . In the given figure two circles intersects at M and N respectively PQ and RS pass through M. R, P, S, Q are joined with N. Prove that PNR=SNQ.

h)

lbOPsf] lrqdf lrqdf AEC = BFD 5 . l;4 ug{ 'xf] xf];\ ;\ . . i) ABC = BCD ii) AB//CD

In the given figure AEC=BFD. Prove that i) ABC = BCD ii) AB//CD

lbOPsf] lrqdf, lrqdf, PQ = PR 5g\ eg] eg] QR//ST x'G5 G5 egL b]vfpg' vfpg'xf] xf];\ ;\ . .

i)

In the given figure PQ=PR. Prove that QR//ST.

lbOPsf] lrqdf lrqdf b'O{O hLjfx? { hLjfx? PQ / RS ljGb' X df nDa x'g]gu/L u] /L sfl6Psf 5g\ . . k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ M

j)

=

In the adjoining figure chords PQ and RS intersects at X. Prove that, k)

=

tn lbOPsf] lrqdf lrqdf ABCD Pp6f rlqmo rt'e{ e 'h { xf] . . olb AB = AC eP BDE sf] cw{ cw{s AD xf] egL egL l;4 ug{ 'xf] xf];\ ;\ .

In the given figure ABCD is a cyclic quadrilateral. If AB=AC. Prove that AD is the bisector of BDE. l)

lbOPsf] lrqdf lrqdf ABC Pp6f lqe'h xf] . . h;df AB = AC 5 . k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . AD = AE In the given figure ABC is a triangle in which AB=AC. Prove that AD=AE.

# Section F

a)

lbOPsf] lrqdf lrqdf PQ//RS eP PTR =

b)

QUS

x'G5 G5 egL k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ M MIn the given figure PQ//RS prove that PTR=QUS.

;Fu} usf] }sf] lrqdf lrqdf ABCD Pp6f ;=r= xf] . . a[Qn] Qn] AB nfO{ E / DC nfO{ F df sf6]sf] sf] 5 5 eg], k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ EFD = ABC

In the adjoining figure ABCD is a parallelogram. The Inscribed circle cuts AB at E and CD at F Prove that EFD=ABC.



c)

lbOPsf] lrqdf lrqdf ABCD rlqmo rt'e{ e 'h { xf] / / AB//DC 5 . l;4 ug{ 'xf] xf];\ ;\ . . (i) AD=BC. (ii) AC=BD.

In the adjoining figure ABCD is a cyclic quadrilateral quadrilateral AB//DC. Prove that (i) AD=BC. (ii) AC=BD.

d)

lbOPsf] lrqdf, lrqdf, ∆ABC ;djfx' lqe' lqe'h 5

. BAC sf] cw{ cw{s AD eP ∆BCE ;dl4jfx' lqe" lqe"h xf] egL egL k|dfl0ft dfl0ft

ug{ 'xf] xf];\ ; . \ . In the given figure, ABC is an inscribed equilateral triangle if AD is the bisector of BAC, prove that BCE is an isosceles triangle.

e)

lbOPsf] lrqdf, lrqdf, b'O{O{ j[ j[Qx? Qx? P / Q ljGb'x?df x?df Ps cfk;df sflf6Psf 5g\ . . P / Q af6 b'O{O /] { /]vfx? vfx? APB / CQD uPsf 5g\

eg] k| k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . . In the given figure, APB and CQD are the straight lines through the points of intersection of two circles. Prove that

(i) AC//BD f)

;Fu} u}sf] sf] lrqdf lrqdf

(ii) CPD=AQB.

Gb| ePsf] ePsf] j[ j[Qdf Qdf AMOXOZ O s]Gb|

eP egL k|dfl0ft dfl0ft ug{ 'xf] xf];\ ; . \ . OAZ=XYO.

In the adjoining figure, O is the center of circle. If AMOXOZ, then prove that OAZ=XYO.

g)

lbOPsf] lrqdf lrqdf DE Aof; 5 . olb BE = CE df AED = (ABC - ACB) x'G5 G5 egL k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . .

In the given figure, DE is a diameter. If arc BE=arc CE, then prove that AED=½ (ABC- ACB) ACB)

h)

lrqdf s]Gb| Gb|ljGb' ljGb'x? x? X / Y ePsf j[Qx?sf] Qx?sf] :kz{ :kz{ljGb' ljGb' P eP k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . . XQ//YR.

In the figure the point of contact of two circles having center X and Y is P. Prove that XQ//YR.

j[Qdf Qdf O s]Gb| Gb|ljGb' ljGb' xf] xf] . . AB / CD b'O{O{ :kz{ :kz{ i) lbOPsf] j[

/]vfx? vfx? x'g\ g\ . . k|dfl0ft dfl0ft

ug{ 'xf] xf];\ ;\ BAC = 2OBC

In the given figure O is the centre of the circle AB and CD are two tangents Prove that BAC = 2OBC.



lrqdf rlqmo rt'e{ e 'h { ABCD sf] Ps Ps e'hf hf BC nfO{ CP ] u/L ljGb' P ;Dd nDafO{Psf] Psf] 5 5 . olb ABC sf] cw{ cw{s CP = AB x'g u/L

j)

BD 5

eg] ∆DBP ;dl4afx' lqe' lqe'h xf] egL egL l;4 ug{ 'xf] xf];\ ;\ . .

In the adjoining figure the side BC of a cyclic quadrilateral ABCD is produced to the point P making CP=AB. If BD bisects ABC, prove that DBP is an isosceles triangle.

k)

rlqmo ii) BXYC \rlqmo

;Fu} usf] }sf] lrqdf lrqdf AB sf] dWoljGb' dWoljGb' X 5 . SAT :kz{/f] /f]vf vf xf] . . ST//XY 5 . k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . . i) AYX = ABC

rt'e{ e 'h { xf] . .

In the adjoining figure X is the midpoint of AB, SAT is tangent & ST//XY. Prove that (i) AYX=ABC (ii) BXYC is a cyclic quadrilateral.

l)

;Fu} usf] }sf] lrqdf lrqdf PT Aof; xf] . . rfk SR = rfk RT eP

k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . . PS//OR.

In the adjoining figure PT is a diameter Arc SR = Arc RT. Prove that PS//OR.

m) lbOPsf] lrqdf lrqdf

j[Qsf Qsf hLjfx?

xo MN / RS af\xo

ljGb'

X df

sfl6Psf 5g\ . . eg] l;4 l;4 ug{ 'xf] xf];\ ;\ . .

MXR=(Arc MR - Arc NS).

In the given figure chords MN and RS intersect at external point X. Prove that MXR = (Arc MR - Arc NS).

# Section G

a)

lrqdf

P / Q qmdzM

rfk

AB /

rfk

AC sf

dWoljGb'x? x? x'g\ g\ eg] eg]

G5 AX = AY x'G5


In the figure P and Q are the mid-points of arc AB and arc AC respectively. Prove that: AX = AY. Hints:

- VII(a) 1. Join PA and QA. 2. PQA = PAB =a 3. CAQ = QPA = b

4. AXY = PAB + QPA = a + b

5. AYX = CAQ + AQP = b + a

6. AXY = AYX b)

lrqdf, hLjfx?

PR / QS k/:k/

7. AX = AY

;dsf]0f 0f kf/Lsg ljGb'

E df

sfl6Psf 5g\ . .

dWo QR sf] dWo

ljGb'

X / XE a9fpbf PS sf] Y df

e]6\ 65 \ eg]

G5 EYPS x'G5


In the figure chords PR and QS of a circle intersect at point E at right angles. X is the mid-point of QR and XE is produced meets PS in Y Prove that EYPS.


Downloaded From:- www.bhawesh.com.np c)

lbOPsf] lrqdf lrqdf

G5 PQ//AB eP AY = BY x'G5


In the given figure, PQ//AB. Prove that AY = BY. d)

lbOPsf] lrqdf lrqdf

G5 PB = QB eP MN//PQ x'G5


In the given figure, PB = QB. Prove that MN//PQ.

olb lbOPsf] lrqdf lrqdf

e)

CMAB /

e'hf

MN//DE eP AMNC Pp6f

rlqmo rt'e{ e '{h /

G5 CNAE x'G5


In the given figure, CMAB and MN//DE, then prove that AMNC is a cyclic quadrilateral and CNAE. Hints: VII(e)

Join A & C (i) CAN = CMN (both are equals to CDE so, AMNC is a Cyclic quadrilateral (ii) AMC = CNA [since both are on the

same arc hence CNAE]

ABCD Pp6f

f)

e]6]6sf s ] f 5g\ . .

rlqmo rt'e{ e 'h { xf], h;df e'hfx? hfx?

cw{sn] sn] DC nfO{ ljGb' ljGb' P df ALB sf] cw{

/

AB nfO{ R df

/

bf AD / BC nDAofpFbf

ljGb' L df / e'hfx? hfx?

cw{sn] sn] BC nfO{ ljGb' ljGb' Q df AMD sf] cw{

/

bf AB / DC nDAofpFbf

AD nfO{ S df

ljGb'

M df

e]6]6]sf sf 5g\ eg] eg] k| k|dfl0ft dfl0ft

ug{ 'xf] xf];\ ; . \ . i) PR / QS k/:k/ ii) PQRS Ps

;dsf]0f 0f x'g]gu/L u] /L sf6\5g\ 5g\ . .

;djfx' rt' rt'e{ e '{h xf] .

ABCD is a cyclic quadrilateral in which the sides AD and BC when produced meet at L and the sides AB and DC, when produced meet at M. AMD meets BC in Q and AD in S. Prove that i) PR and QS intersect at right angles Hints VII(f)

ii) PQRS is a rhombus.

1. In LAR & LPC i. ALR =PLC ii. LAR = LCP 2. LAR LPC [by A.A case] 3. ARL = DPR 4. MPR = MRP

5. MP = MR

7. Similarly SN = NQ & LNSQ

g) A / B df

:kz{ /] /]vf vf

QR ;Fu

6. PN P N = RN & MNPR 8. PQRS is a rhombus

k|ltR5] ltR5]lbt lbt b'O{O j[ { j[Qx?df Qx?df Pp6f j[Qdf Qdf s'g}g ljGb' } ljGb'

bf P 5 PA / PB nDAofpFbf

csf{ ] j[ j[Qdf Qdf

Q / R

df e]6\ 6\5 eg]

df P ljGb'df

lvr]sf] sf]

;dfgfGt/ x'G5 G5 egL k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . .

Two circles intersect at A and B. P is any point on one circle PA and PB produced & meet other circle at Q and R, prove that tangent drawn on a point P is parallel to QR.

h)

lbOPsf] lrqdf lrqdf

Qsf] s] s]Gb| Gb|ljGb' ljGb' CE :kz{/]/vf v ] f O j[Qsf]

/

ljGb' 5 5 D :kz{ljGb'

eg] k| k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . .

In the given figure, O is the centre of the circle. CE is a tangent and D is the point of contact then prove that: BOD = 2(AFD-ACD). Hints :

VII(h) Join BD & DA


Downloaded From:- www.bhawesh.com.np 1. Let AFD = ABD = a, BAD = PDC = b & BCD = c 2. BOD = 2BAD = 2b

3. a = b + c

4. BOD = 2(a-c)

b = a - c 2b = 2a-2c

or, BOD = 2(AFD-ACD)

# Section H a) lrqdf A, B, C, D, P, Q, R, S kl/lwsf

ljGb'x? x? eP

P + Q + R + S = 540

x'G5 G5 egL k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . .

0

In the figure A, B, C, D, P, Q, R, S are the points at the circumference then prove that: P + Q + R + S = 540 Hints:

0

0

VIII(a) Draw BD then opposite angles of cyclic quadrilateral is 180 so. 0

0

0

0

P + ADB = 180 , S + ABD = 180 , R + DBC = 180 and Q + CDB = 180 Adding all of them (P + Q +S) + [(ADB + CBD) +(ABD + DBC)] = 720 or,

0

0

(P + Q + R + S) + (ADC + ABC) = 720 0

0

or, (P + Q + R + S) + 180 = 720

lrqdf

b)

0

P + Q + R + S = 540

lqe'h ABC ;dafx' lqe'

xf] . .

D / E qmdzM

rfk AB / rfk

AC sf

dWolaGb'x? x? eP k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ M M

3XY = DE

In the adjoining figure, ABC is an equilateral triangle. D and E are the midpoints of arc AB and arc AC respectively. Prove that : 3XY = DE. Hints:

VIII(b)

(1) = = = =[Being ABC equilateral from given] (2) DAX = ADX [Both are angles at the circumference and = ] (3) DX = AX [From (ii)]

(4) Similarly AY = YE[Same as above]

(5) AX = AY = XY [Sides of an equilateral triangle] (6) DX = YE[From (iii), (iv) and (v)] (7) DE = DX + XY + YE = XY + XY + XY = 3XY c) ABC Pp6f

;dl4afx' lqe' lqe'h j[QcGtu{ QcGtu{t 5 .

B / C sf

cw{sx? sx? kl/lwsf laGb'x? x?

X / Y df

qmdzM e]6 ePsf 5g\ . . ax'e' e'h

BXAYC sf

rf/cf]6f 6f

e'hfx? hfx? a/fa/ x'G5g\ G5g\ egL egL k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . . ABC is an isosceles triangle inscribed in the circle, the bisectors of B and C meet the circumference at X and Y respectively. Shoe that the polygon BXAYC must have four of its sides equal. Hints:

VIII(c) 1) ACY = BCX [Being CX is a bisector bisector of ACB]

2) ABY = YBC [Being BY is a bisector of ABC] 3) arc AX = arc BX = arc AY = arc YX [Corresponding areas of (i) and (ii)] 4) AX = BX = AY = YC [Corresponding segment of (iii)]

d)

lbOPsf] lrqdf lrqdf

BAC sf] cw{ cw{s AP 5

eg]

EF / BC ;dfgfGt/

5g\ egL egL k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . .

In the given figure, AP is bisector of BAC. Prove that EF and BC are parallel. Hints:VIII(d)

(1) Join AD then being ABP = EAF and BAP = EDF we get EAF = EDF and AEFD is a cyclic quad.

(2) EFA = EDA and EDA = ACB gives EFA = ACB and EF//BC e) ABCD Pp6f

ju{ / /

AEF Pp6f

;dafx' lqe' lqe'h Pp6} j[ j[Qdf Qdf cGtu{t 5g\ EF//BD x'G5 G5 elg k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . .



ABCD is a square and AEF is an equilateral triangle inscribed in the same circle. Prove that EF//DB. Hints:VIII(g)

(1) Being AD = AB:AD = AB

(2) AE = AF:ADE = ABF

(3) DE = BF(subtracting (i) from (ii) hence BD/EF.)

f) ABC df, x, y z qmdzM BC, AC / AB sf

dWoljGb'x? x? x'g\ g\ olb olb

x? AWBC eP W,X,Y, Z ljGb'x?

rqmLo x'g egL k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . .

In ABC, X, Y and Z are the mid-points of BC, AC and AB respectively. If AWBC, prove that W, X, Y and Z are co cyclic. Hints:VIII(f) Join WZ

(1) AZ = BZ = WZ [Being ABW rt. angled ] (2) ZBXY is a parm. so B = ZYX and (3) Being BZ = ZW, B = ZWB hence ZWB = XYZ and W, X, Y, Z are con-cyclic. g)

;Fu} u}sf] sf] lrqdf lrqdf

AB = BC, AC//XY, CX, AY, XD / YD x?

;Lwf/]vfx? vfx? eP k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ ACYX e '{h xf] . . ACYX Pp6f rqmLo rt'e{

In the adjoining figure, AB = BC, AC//XY, CX, AY, XD and YD are straight lines. Prove that: ACYX is a cyclic quadrilateral

Hints:

VIII(g) (1) AB = BC, BAC = BCA and

being AC//XY, BAC = BYX (2) Combing all we get BCA = BYX

h)

lrqdf

ABC Pp6f

ACYX is a cyclic quadrilateral

cGtu{t lqe'h xf] . . h;df

PMAB,PNAC / PRBC 5

eg]

MNR Pp6f

;Lwf/]vf vf x'G5 G5 egL k|dfl0ft dfl0ft ug{ 'xf] xf];\ ;\ . .

ABC is an inscribed triangle, PMAB,PNAC and PRBC. Prove that MNR is a straight line. Hints:VIII(h)

Join PA and PC

1) Show ANPM is a cyclic quad 2) Show PNRC is a cyclic quad. 3) PCB = PAM[Being ABCP is a cyclic quad.] 4) PCR = PAM 5) PNM = PAM 6) PCR = PNM 0

7) PNR + PCR = 180 [Being the opposite angles of a cyclic quad.] 0

8) PNR + PNM = 180 [from 6 & 7] 9) RNM is a straight line [Form (8)]


SLC _ Compulsory Math _ Geometric Deduction

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