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 9^{th} 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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# 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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Given:

ABCD is a parallelogram

AD = CQ

To prove:

ar (â–³BPC) = ar (â–³DPQ)

Proof:

In â–³ADP and â–³QCP,

âˆ APD = âˆ QPC [Vertically Opposite Angles]

âˆ ADP = âˆ QCP [Alternate Angles]

AD = CQ [given]

, 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)