Question

$ABCD$ is a parallelogram, points $P$ and $Q$ on $BC$ trisects $BC$. Prove that :-

$ar\left(△APQ\right)=ar\left(△DPQ\right)=\frac{1}{6}ar\left(parallelogramABCD\right)$

Answer 1

Through P and Q, draw PR and QS parallel to AB .Now,PQRS is a parallelogram and its base PQ =1/3 BC.

Since △ APQ and △DPQ are on the same base PQ, and between the same parallels AD and BC.

$ar(△APQ)=ar(△DPQ)⟹\left(1\right)$

Since △APQ and parallelogram PQSR are on the same base PQ, and between same parallels PQ and AD.

$ar(△APQ)=\frac{1}{2}ar\left(||gmPQRS\right)⟹\left(2\right)$

$\frac{ar\left(||gmPQRS\right)}{ar\left(||gmABCD\right)}=\frac{BC\ast height}{PQ\ast height}=\frac{3PQ}{PQ}$

$ar\left(||gmPQRS\right)=\frac{1}{3}ar\left(||gmABCD\right)⟹\left(3\right)$

Using (2) and (3),

$ar(△APQ)=\frac{1}{2}ar\left(||gmPQRS\right)=\frac{1}{2}\ast \frac{1}{3}ar\left(||gmABCD\right)$

$ar(△APQ)=ar(△DPQ)=\frac{1}{6}ar\left(||gmABCD\right)$