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.]

Solution:

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)