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In Figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersects DC at P, show that ar (BPC) = ar (DPQ).

  

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In Figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersects DC at P, show that ar (BPC) = ar (DPQ).

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Given,

ABCD is a parallelogram

AD = CQ

To prove:

ar (△BPC) = ar (△DPQ)

Proof:

In △ADP and △QCP,

∠APD = ∠QPC [Vertically Opposite Angles]

∠ADP = ∠QCP [Alternate Angles]

AD = CQ [given]

△ABO ≅ △ACD [AAS congruency]

DP = CP [CPCT]

In △CDQ, QP is median. [Since, DP = CP]

Since, median of a triangle divides it into two parts of equal areas.

ar(△DPQ) = ar(△QPC) —(i)

In △PBQ, PC is median.

[Since, AD = CQ and AD = BC ⇒ BC = QC]

Since, median of a triangle divides it into two parts of equal areas.

ar(△QPC) = ar(△BPC) —(ii)

From the equation (i) and (ii), we get

ar(△BPC) = ar(△DPQ)

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