Question

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

Answer 1

Since $△APB$ and parallelogram ABCD are on the same base AB and between the same parallels.

Therefore,

$ar\left(△APB\right)=\frac{1}{2}ar\left({\parallel }^{gm}ABCD\right)$ .......(1)

Similarly,

$ar\left(△BQC\right)=\frac{1}{2}ar\left({\parallel }^{gm}ABCD\right)$ ......(2)

From 1 and 2, we get,

$ar\left(△APB\right)=ar\left(△BQC\right)$