Question

In Fig., $ 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.\left(BPC\right)$$ =$$ ar.\left(DPQ\right)$. [Hint : Join $ AC$.]

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Answer 1

Step 1: Simplify the given information by construction

Given diagram $ABCD$ is a parallelogram, where $AD=CQ$

Join $AC$

Step 2: Prove the area of given triangles equal

We need to prove, $ar.\left(BPC\right)$$=$$ar.\left(DPQ\right)$

$ACQD$ will be parallelogram ($\therefore AD=CQ,AD\parallel CQ$)

In $∆APC$ and $∆QPD$

$AC=QD$ [opposite side of parallelogram]

$\angle PCA=\angle PAC$ [Altitude interior angles]

$\therefore ∆APC\cong ∆QPD$ [From A.S.A congruence]

$ar.\left(APC\right)=ar.\left(BPC\right)$ [because both lie on the same base $PC$ and between same $\parallel$lines $PC$ and $AB$]

From above equations

$ar.\left(BPC\right)$$=$$ar.\left(DPQ\right)$.

Hence proved.