Question

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

Answer 1

Solution :-

ABCD is a parallelogram so $AB\parallel CD$ & $AD\parallel BC$

$△APB$ and parallelogram ABCD

lie on the same base AB

and are between the

same parallel lines AB and DC,

$△QBC$ and parallelogram

ABCD lie on the same

base BC and are

between the same

parallel lines AD and BC.

Area of triangle is half

of parallelogram if they

have the same base and parallels

so

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

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

From $e{q}^n$ (1) & (2)

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

$H.P$