Question

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

Answer 1

Answer:

Refer image, given ABCD is a parallelogram

AP∥CR

CP=PD

Since, △APD∼△RCD(AAA)

⇒

DR

DA

=

DC

DP

(CPCT)

⇒

DR

DA

=

2

1

(AsPisthemidpointofCD)

⇒2DA=DR

⇒2DA=DA+AR

⇒DA=AR

Hence Proved DA=AR

Again as, ∠APD=∠QCD&∠BQC=∠APD(AP∥QC),

⇒∠APD=∠QCD=∠BQC

&

∠RDC=∠CBQ(Oppositeof∠is∥gm)

&

∠BCQ=∠CRD(Propertiesof∥gm)

⇒△BQC∼△DCR(AAA)

Now,

As,

⇒△APD≅△CBQ

PD=BQ

or

2

CD

=BQ(AsPisthemidpointofCD)

or

2

AB

=BQ(In∥gmAB=CD)

⇒ Q is the mid-point of AB

Again,

as,

△BQC∼△DCR

CQ

BQ

=

CR

CD

⇒2CQ=CR(AsCD=2BQ)

⇒2CQ=CQ+QR

⇒CQ=QR

Hence Proved CQ=QR

Answer 2

Answer:

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