1.

P is the mid-point of the 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»

Given: ABCD is a parallelogram in which P is the mid-point of side CD.

To prove: DA = AR and CQ = QR

Proof: ABCD is a parallelogram.

BC = AD and BC || AD

Also, DC = AB and DC || AB

Since, P is mid-point of DC

DP = PC = \(\frac { 1 }{ 2 }\) DC

Now, QC || AP and PC || AQ.

Hence, APCQ is a parallelogram.

AQ = PC = \(\frac { 1 }{ 2 }\) DC = \(\frac { 1 }{ 2 }\) AB

= BQ [∵ DC = AB]

Now in ∆AQR and ∆BQC

AQ = BQ

∠AQR = ∠BQC (vertically opposite angles)

and ∠ARQ = ∠BCQ (alternate interior angles)

∆AQR = ∆BQC (by AAS congruence rule)

So, AR = BC (by c.p.c.t)

But, BC = DA

AR = DA

Also, CQ = QR (by c.p.c.t)

Hence proved.


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