Question

$P$ and $Q$ are any two points lying on the sides $DC$ and $AD$ respectively of a parallelogram $ABCD$. Show that area $\left(APB\right)=area\left(BQC\right)$.

Answer 1

$ABCD$ is a parallelogram.

So, $AB\parallel CD$ and $AD\parallel BC$

$△APB$ and parallelogram $ABCD$ lie on the same base

$AB$ and are between the same parallel lines $AB$ and $DC,$.

We know that, area of triangle is half of parallelogram if they have the same base and parallels.

$\therefore$ $Area\left(△APB\right)=\frac{1}{2}Area\left(ABCD\right)$ ---- ( 1 )

$△QBC$ and parallelogram $ABCD$ lie on the same base

$BC$ and are between the same parallel lines $AD$ and $BC,$.

We know that, area of triangle is half of parallelogram if they have the same base and parallels.

$\therefore$ $Area\left(△BQC\right)=\frac{1}{2}Area\left(ABCD\right)$ ---- ( 2 )

From ( 1 ) and ( 2 ),

$\Rightarrow$ $Area\left(△APB\right)=Area\left(△BQC\right)$ --- Hence proved