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Question

P is mid point of the side CD of a parallelogram ABCD. A line through C parallel to PA intersect AB at Q and DA produced at R. Prove that DA=AR and CQ=QR.

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Solution

Refer attached image, given ABCD is a parallelogram

APCR

CP=PD


Since, APDRCD(AAA)

DADR=DPDC(CPCT)

DADR=12(AsPisthemidpointofCD)

2DA=DR

2DA=DA+AR

DA=AR


Hence Proved DA=AR


Again as, APD=QCD&BQC=APD(APQC),


APD=QCD=BQC

&

RDC=CBQ(Oppositeofisgm)

&

BCQ=CRD(Propertiesofgm)


BQCDCR(AAA)


Now,

As,

APDCBQ

PD=BQ

or CD2=BQ(AsPisthemidpointofCD)

or AB2=BQ(IngmAB=CD)

Q is the mid-point of AB


Again,

as,

BQCDCR

BQCQ=CDCR


2CQ=CR(AsCD=2BQ)

2CQ=CQ+QR

CQ=QR


Hence Proved CQ=QR


1026546_1081204_ans_7812f72c25bc461d8b5805e35f139be7.png

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