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# In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ). [Hint : Join AC.] Q.4

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Hello sir i want to know the best solution of the question from exercise 9.4 of math of Areas of Parallelograms and Triangles chapter of class 9th give me the best and easy for solving this question how i solve it of question no. 4 In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ). [Hint : Join AC.

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1. Given:

ABCD is a parallelogram

To prove:

ar (â–³BPC) = ar (â–³DPQ)

Proof:

âˆ APD = âˆ QPC [Vertically Opposite Angles]

âˆ ADP = âˆ QCP [Alternate Angles]

, â–³ABO â‰… â–³ACD [AAS congruency]

, DP = CP [CPCT]

In â–³CDQ, QP is median. [Since, DP = CP]

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

, ar(â–³DPQ) = ar(â–³QPC) â€”(i)

In â–³PBQ, PC is median. [Since, AD = CQ and AD = BC â‡’ BC = QC]

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

, ar(â–³QPC) = ar(â–³BPC) â€”(ii)

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

ar(â–³BPC) = ar(â–³DPQ)

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