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Mathematics Practice Questions
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# Section A
a) lbPsf] lrqdf ABCD df CD Pp7f ;r xf] / CD sf] s'g} ljGb' P 5 eg] APD
+ BCP = APB x'G5 egL k|dfl0ft ug{'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{'h xf] . AB sf] s'g} ljGb' P / AD sf] s'g} ljGb' Q
5 eg] QBC = APD + CPB x'G5 elg k|dfl0ft ug{'xf];\ .
ABCD is a parallelogram. P is any point in AB and Q is any point on
AD. Prove that. QBC = APD + CPB.
c) lbOPsf] lrqdf ABCD Pp6f ;=r= xf] h;df ljs0f{ BD df kg{] s'g} ljGb' X 5 .
k|dfl0ft ug{'xf];\ .
1
AXB = 2 rt'e{'h ABCX.
In the given figure, ABCD is a parallelogram is which X is any point on the diagonal
1
BD. Prove that : AXB = 2 quadrilateral ABCX.
d) ;Fu}sf] lrqdf PQRSPp6f ;dfgfGt/ rt'e{'h xf] . ljs0f{ PR sf] s'g} ljGb' M
;Fu Q / S hf]l8Psf 5g\ . l;4 ug{'xf];\ .
PQM sf] If]qkmn = PSM sf] If]qkmn
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{'h ABCD df AO = Co eP 2ABD = ABCD x'G5 egL k|dfl0ft
ug{'xf];\ .
In quadrilateral ABCD, AO = Co, prove that 2ABD = ABCD.
f) ABC sf] zLif{ljGb' A / BC sf] Pp6f ljGb' D hf]l8Psf] 5 / AD sf]
dWoljGb' E 5 eg] ABC = 2EBC x'G5 elg k|dfl0ft ug{'xf];\ .
The vertex A of ABC is joined to a point D on the side BC. The
midpoint of AD is E, Prove that ABC = 2EBC.
# Section B
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a) ;Fu]sf] lrqdf ABCD Pp6f ;dfgfGt/ rt'e{'h xf] . CD df s'g} Pp6f ljGb' N 5 . AN
n] BC nfO{ M ljGb'df 5'g] u/L a9fOPsf] 5 eg] BNC sf] If]qkmn = DNM
sf] If]qkmn x'G5 elg k|dfl0ft ug{'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 ABCD Pp6f ;=r= xf] . olb DC = CQ eP PQC = BPQ x'G5
egL k|dfl0ft ug{'xf];\ .
In the given figure, ABCD is a parallelogram IF DC = CQ, prove that
PQC = BPQ.
c) ;=r= ABCD sf] e'hf AD df s'g} ljGb' F 5 BF / CD nfO{ a9fP/ E ljGb'df
ldnfOPsf] 5 . k|dfl0ft ug{'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.
d) ABC sf] e'hf BC ;Fu ;dfgfGt/ /]vf XY 5 . BE//AC / CF//AB n] XY nfO{ E
/ F df e]6\5 eg] ABE = ACF x'G5 elg k|dflf0ft ug{'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
e) ABC df AB sf] dWoljGb' D / BC sf] s'g} ljGb' P 5 olb . CQ//DP eP
2BPQ = ABC x'G5 elg k|dfl0ft ug{'xf];\ .
In ABC, D is the mid-point of AB and P is any point on BC. IF CQ//DP, prove that:
2BPQ = ABC.
f) lbOPsf] lqe'h ABC df, BE / CD b'o{ dlWosfx¿ ljGb' O df k|lt5]lbt
5g\ . l;4 ug{'xf];\ . BOC sf] If]qkmn = rt'e{'h ADOE sf] If]qkmn
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¿ AB, BC, / CA sf dWoljGb'x¿ qmdzM P, Q, R 5g\ eg]
k|dfl0ft ug{'xf];\ .
P, Q and R are the midpoints of sides AB, BC and CA respectively
of ABC, prove that,
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1
APR = BPQ = CQR = PQR = 4 ABC.
h) lbOPsf] lrqdf AP, BP / CP sf dWoljGb'x¿ qmdzM L, M / N x'g\ eg]
ABC = 4LMN x'G5 elg k|dfl0ft ug{'xf];\ .
In the given figure, L, M and N are the mid-points of AP, BP and CP respectively,
prove that ABC = 4LMN.
# Section C
a) rt'e{'h ABCD sf] lzif{ljGb' C af6 ljs0f{ DB ;Fu ;dfgfGt/ x'g] u/L lvrLPsf]
/]vfn] AB nfO{ nDafOPsf] /]vfsf] ljGb' E df 5f]Psf] 5 . l;4 ug{'xf];\ rt'e{'h
ABCD sf] If]qkmn = DAE sf] If]qkmn
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 PQRS ;dnDa rt'e{'h xf], h;df PQ//MN//SR 5g\ eg] l;4
ug{'xf];\ .
PSN sf] If]qkmn = QRM sf] If]qkmn
In the given figure, PQRS is a trapezium in which PQ//MN//SR. Prove that area of
PSN = area QRM.
# Section D
a) rlqmo rt'e{'h ABCD df AP, BP, CR / DR qmdzM A, B, C / D sf
cw{sx?åf/f ag]sf] rt'e{'h QPRS klg rlqmo rt'e{'h x'G5 elg k|dfl0ft
ug{'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 j[Qsf] s]Gb| ljGb' O, Jof; AB /
ODA x'G5 elg l;4
DOAB eP AEC =
ug{'xf];\ .
In the adjoining figure O is the center of a circle. AB is a diameter and DOAB. Prove
AEC=ODA.
c) lbOPsf] lrqdf AD//BC 5 eg] AYC = BXD
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x'G5 egL k|dfl0ft ug{'xf];\ .
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Given figure is a circle where AD//BC, Prove that AYC=BXD.
d)
lbOPsf] lrqdf AC = BC / ABCD Pp6f rlqmo rt'e{'h xf] . eg] CD n] BDE nfO{ cfwf
ub{5 egL k|dfl0ft ug{'xf];\ .
In the given figure, AC=BC and ABCD is a cyclic quadrilateral prove that DC bisects
BDE.
e) b'O{ hLjfx? AB / CD k/:k/ nDa 5g\ AOD +
BOC = 1800 x'G5 elg
k|dfl0ft ug{'xf];\ .
Two chords AB and CD are perpendicular to each other. Prove
AOD+BOC=1800.
f) lbOPsf] lrq Pp6f j[Qsf] xf] . h;df PMS = QNR eg] PQ//RS x'G5 egL
k|dfl0ft ug{'xf];\ .
Given figure is a circle and PMS = QNR Prove PQ//RS.
g) lbOPsf] lrqdf s]Gb| ljGb' O ePsf] j[Qsf b'O{ Aof;x? AB / CD x'g\ . olb hLjf
CE Aof; AB ;Fu
;dfgfGt/ / BOC clws sf]0f x'g\ eg] rfk DBE
sf] dWo
ljGb' B x'G5 egL l;4 ug{'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 ABCD Pp6f rlqmo rt'e{'h xf] . olb BC = DE / CA n] BCD
nfO{ ;dlåljefhg ub{5 eg] ∆ACE ;dlåjfx' lqe'h x'G5 egL k|dfl0ft ug{'xf];\ .
In the given figure, ABCD is a cyclic quadrilateral. If BC=DE and CA
bisects BCD, prove that ACE is an isosceles triangle.
i) lbOPsf] j[Qdf AB Aof; xf] . OC / OD Aof;fw{x?
x'g\, hxfF rfk BC = rfk
CD 5 eg] AD//OC x'G5 egL k|dfl0ft ug{'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
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a) lbOPsf] j[Qdf AB Aof;, OC / OD Aof;fy{x? x'g\, hxfF BOC =COD
x'g\ eg] AD//OC x'G5 egL l;4 ug{'xf];\ .
In the adjoining figure, AB is the diameter. OC and OD are radii where
BOC=COD, prove that AD//OC.
b) lbOPsf] lrqdf BC, DBE sf] cw{s xf] . ABCD rlqmo rt'e{'h xf] eg] k|dfl0ft
ug{'xf];\ AC = DC
In the given figure BC is a bisector of DBE. ABCD is a cyclic
quadrilateral Prove that AC=DC.
c) O / P s]Gb|ljGb' ePsf b'O{ j[Qx? A / B df k|ltR5]lbt ePsf 5g\ . A / B tyf O
/ P hf]8\bf C df sf6LPsf 5g\ k|dfl0ft ug{'xf];\ .
i) AC = BC
ii) OCA = 900
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=900.
d) O s]Gb|ljGb' ePsf] j[Q XPY / MQN nfO{ /]vf XY
sf6]sf] 5 k|dfl0ft ug{'xf];\ .
n] X / Y tyf M / N df
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}sf] lrqdf ljGb' A af6 AB / AC :kz{
ljGb'
/]vfx? lvlrPsf 5g\ . :kz{ /]vf DE
F df :kz{ ePsf 5g\ eg] k|dfl0ft ug{'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? j[Qsf]
:kz{ ljGb'df :kz{
ePsf
3]/fsf] qmdzM P,Q,R / S
5g\ k|dfl0ft ug{'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 b'O{ j[Qx? M / N df k|ltR5]lbt 5g\ . PQ / N df ljGb' M af6
kf/ ePsf 5g\ . R,P,S / Q qmdzM ljGb' N ;Fu hf]l8Psf 5g\ eg] l;4
ug{'xf];\ .
In the given figure two circles intersects at M and N respectively PQ
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and RS pass through M. R, P, S, Q are joined with N. Prove that PNR=SNQ.
h) lbOPsf] lrqdf AEC = BFD 5 . l;4 ug{'xf];\ . i) ABC = BCD ii) AB//CD
In the given figure AEC=BFD. Prove that i) ABC = BCD ii)
AB//CD
i) lbOPsf] lrqdf, PQ = PR 5g\ eg] QR//ST x'G5 egL b]vfpg'xf];\ .
In the given figure PQ=PR. Prove that QR//ST.
j) lbOPsf] lrqdf b'O{ hLjfx? PQ / RS ljGb' X df nDa x'g]u/L sfl6Psf 5g\ . k|dfl0ft
ug{'xf];\ M
PS - QS = PR - QR
In the adjoining figure chords PQ and RS intersects at X. Prove that, PS - QS =
PR - QR
k) tn lbOPsf] lrqdf ABCD Pp6f rlqmo rt'e{'h xf] . olb AB = AC eP BDE sf]
cw{s AD xf] egL l;4 ug{'xf];\ .
In the given figure ABCD is a cyclic quadrilateral. If AB=AC. Prove that
AD is the bisector of BDE.
l) lbOPsf] lrqdf ABC Pp6f lqe'h xf] . h;df AB = AC 5 . k|dfl0ft ug{'xf];\ . AD =
AE
In the given figure ABC is a triangle in which AB=AC. Prove that AD=AE.
# Section F
a) lbOPsf] lrqdf PQ//RS eP PTR =
QUS
k|dfl0ft ug{'xf];\ MIn the given figure PQ//RS prove
x'G5 egL
that
PTR=QUS.
b) ;Fu}sf] lrqdf ABCD Pp6f ;=r= xf] . a[Qn] AB nfO{ E /
DC nfO{ F df
sf6]sf] 5 eg], k|dfl0ft ug{'xf];\ EFD = ABC
In the adjoining figure ABCD is a parallelogram. The
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cuts AB at E and CD at F Prove that EFD=ABC.
c) lbOPsf] lrqdf ABCD rlqmo rt'e{'h xf] / AB//DC 5 . l;4
ug{'xf];\ . (i)
AD=BC. (ii) AC=BD.
In the adjoining figure ABCD is a cyclic quadrilateral
AB//DC. Prove
that (i) AD=BC. (ii) AC=BD.
d) lbOPsf] lrqdf, ∆ABC ;djfx' lqe'h 5
∆BCE ;dl4jfx' lqe"h xf] egL k|dfl0ft
. BAC sf] cw{s AD eP
ug{'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, b'O{ j[Qx? P / Q ljGb'x?df Ps cfk;df sflf6Psf 5g\ . P / Q
af6 b'O{ /]vfx? APB / CQD uPsf 5g\ eg] k|dfl0ft ug{'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
(ii) CPD=AQB.
f) ;Fu}sf] lrqdf O s]Gb| ePsf] j[Qdf AMOXOZ eP egL k|dfl0ft ug{'xf];\ .
OAZ=XYO.
In the adjoining figure, O is the center of circle. If AMOXOZ, then
prove that OAZ=XYO.
1
g) lbOPsf] lrqdf DE Aof; 5 . olb BE = CE df AED = 2 (ABC - ACB)
x'G5 egL k|dfl0ft ug{'xf];\ .
In the given figure, DE is a diameter. If arc BE=arc CE, then prove that
AED=½ (ABC- ACB)
h) lrqdf s]Gb|ljGb'x? X / Y ePsf j[Qx?sf] :kz{ljGb' P eP k|dfl0ft ug{'xf];\ .
XQ//YR.
In the figure the point of contact of two circles having center X and Y
is P. Prove that XQ//YR.
i) lbOPsf] j[Qdf O s]Gb|ljGb' xf] . AB / CD b'O{ :kz{
/]vfx? x'g\ . k|dfl0ft
ug{'xf];\ BAC = 2OBC
In the given figure O is the centre of the circle AB and CD are two
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tangents Prove that BAC = 2OBC.
j) lrqdf rlqmo rt'e{'h ABCD sf] Ps e'hf BC nfO{ CP = AB x'g] u/L ljGb' P
;Dd nDafO{Psf] 5 . olb ABC sf] cw{s BD 5 eg] ∆DBP ;dl4afx' lqe'h xf]
egL l;4 ug{'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) ;Fu}sf] lrqdf AB sf] dWoljGb' X 5 . SAT :kz{/f]vf xf] . ST//XY 5 .
k|dfl0ft ug{'xf];\ . i) AYX = ABC
ii) BXYC \rlqmo rt'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}sf] lrqdf PT Aof; xf] . rfk SR = rfk RT eP k|dfl0ft ug{'xf];\ . PS//OR.
In the adjoining figure PT is a diameter Arc SR = Arc RT. Prove that
PS//OR.
m)
lbOPsf] lrqdf j[Qsf hLjfx? MN / RS af\xo ljGb' X df sfl6Psf 5g\ .
1
eg] l;4 ug{'xf];\ . MXR=2 (Arc MR - Arc NS).
In the given figure chords MN and RS intersect at external point
1
X. Prove that MXR = 2 (Arc MR - Arc NS).
# Section G
a) lrqdf P / Q qmdzM rfk AB / rfk AC sf dWoljGb'x? x'g\ eg] AX = AY x'G5
egL k|dfl0ft ug{'xf];\ .
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
6.  AXY = AYX
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5. AYX = CAQ + AQP = b + a
7. AX = AY
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b) lrqdf, hLjfx? PR / QS k/:k/ ;dsf]0f kf/Lsg ljGb' E df sfl6Psf 5g\ . QR sf] dWo ljGb' X / XE
a9fpbf PS sf] Y df e]6\5 eg] EYPS x'G5 egL k|dfl0ft ug{'xf];\ .
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
EYPS.
c) lbOPsf] lrqdf PQ//AB eP AY = BY x'G5 egL k|dfl0ft ug{'xf];\ .
In the given figure, PQ//AB. Prove that AY = BY.
d) lbOPsf] lrqdf PB = QB eP MN//PQ x'G5 egL k|dfl0ft
ug{'xf];\ .
In the given figure, PB = QB. Prove that MN//PQ.
e) olb lbOPsf] lrqdf CMAB / e'hf MN//DE eP AMNC Pp6f rlqmo rt'e{'h /
CNAE x'G5 egL k|dfl0ft ug{'xf];\ .
In the given figure, CMAB and MN//DE, then prove that AMNC is a
cyclic quadrilateral and CNAE.
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
CNAE]
f) ABCD Pp6f rlqmo rt'e{'h xf], h;df e'hfx? AD / BC nDAofpFbf ljGb'
L df / e'hfx? AB / DC nDAofpFbf ljGb' M df e]6]sf 5g\ . ALB sf]
cw{sn] DC nfO{ ljGb' P df / AB nfO{ R df / AMD sf] cw{sn] BC
nfO{ ljGb' Q df / AD nfO{ S df e]6]sf 5g\ eg] k|dfl0ft ug{'xf];\ .
i) PR / QS k/:k/ ;dsf]0f x'g]u/L sf6\5g\ .
ii) PQRS Ps ;djfx' rt'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
ii) PQRS is a rhombus.
Hints VII(f) 1. In LAR & LPC i. ALR =PLC ii. LAR = LCP
2. LAR  LPC [by A.A case]
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3. ARL = DPR
4. MPR = MRP
5. MP = MR
6. PN = RN & MNPR
8.  PQRS is a rhombus
7. Similarly SN = NQ & LNSQ
g) A / B df k|ltR5]lbt b'O{ j[Qx?df Pp6f j[Qdf s'g} ljGb' P 5 PA / PB
nDAofpFbf csf{] j[Qdf Q / R df e]6\5 eg] P ljGb'df lvr]sf] :kz{ /]vf QR ;Fu
;dfgfGt/ x'G5 egL k|dfl0ft ug{'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 O j[Qsf] s]Gb|ljGb' CE :kz{/]vf / D :kz{ljGb' 5 eg] k|dfl0ft
ug{'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
1. Let AFD = ABD = a, BAD = PDC = b & BCD = c
2. BOD = 2BAD = 2b
3. a = b + c
 b = a - c  2b = 2a-2c
4. BOD = 2(a-c)
or, BOD = 2(AFD-ACD)
# Section H
a)
lrqdf A, B, C, D, P, Q, R, S kl/lwsf ljGb'x? eP P + Q + R + S = 5400 x'G5
egL k|dfl0ft ug{'xf];\ .
In the figure A, B, C, D, P, Q, R, S are the points at the circumference then prove
that: P + Q + R + S = 5400
Hints: VIII(a) Draw BD then opposite angles of cyclic quadrilateral is 180 0 so.
P + ADB = 1800, S + ABD = 1800, R + DBC = 1800 and Q + CDB =
1800
Adding all of them
(P + Q +S) + [(ADB + CBD) +(ABD + DBC)] = 7200
or, (P + Q + R + S) + (ADC + ABC) = 7200
or, (P + Q + R + S) + 1800 = 7200
Math Practice Questions
Page 10 of 13
 P + Q + R + S = 5400
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Mathematics Practice Questions
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b) lrqdf ABC ;dafx' lqe'h xf] . D / E qmdzM rfk AB / rfk AC sf dWolaGb'x? eP
k|dfl0ft ug{'xf];\ 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 
(1) AD = BD = AE = CE =2 BC [Being ABC equilateral from given]


(2) DAX = ADX [Both are angles at the circumference and BD = AE ]
(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'h j[QcGtu{t 5 . B / C sf cw{sx? kl/lwsf laGb'x? X
/ Y df qmdzM e]6 ePsf 5g\ . ax'e'h BXAYC sf rf/cf]6f e'hfx? a/fa/ x'G5g\
egL k|dfl0ft ug{'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 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 BAC sf] cw{s AP 5 eg] EF / BC ;dfgfGt/ 5g\ egL k|dfl0ft
ug{'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.
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Page 11 of 13
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(2) EFA = EDA and EDA = ACB gives EFA = ACB and EF//BC
e) ABCD Pp6f ju{ / AEF Pp6f ;dafx' lqe'h Pp6} j[Qdf cGtu{t 5g\ EF//BD x'G5 elg k|dfl0ft
ug{'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'g\ olb AWBC eP W,X,Y, Z
ljGb'x? rqmLo x'g egL k|dfl0ft ug{'xf];\ .
In ABC, X, Y and Z are the mid-points of BC, AC and AB
respectively. If AWBC, 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 concyclic.
g) ;Fu}sf] lrqdf AB = BC, AC//XY, CX, AY, XD / YD x? ;Lwf/]vfx? eP k|dfl0ft ug{'xf];\
ACYX Pp6f rqmLo rt'e{'h xf] .
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
 ACYX is a cyclic
quadrilateral
h) lrqdf ABC Pp6f cGtu{t lqe'h xf] . h;df PMAB,PNAC / PRBC 5 eg] MNR Pp6f
;Lwf/]vf x'G5 egL k|dfl0ft ug{'xf];\ .
ABC is an inscribed triangle, PMAB,PNAC and PRBC. Prove that MNR is a
straight line.
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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
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Page 13 of 13
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