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

Parallelogram ABCD and triangle PAB are on the same base AB, and we know that if a triangle and a parallelogram lie on the same base and between two parallel lines then the area of the triangle will be half the area of parallelogram, i.e.

$ar\left(PAB\right)=ar\left(ABCD\right)/\mathrm{2........}\left(i\right)$

also, parallelogram ABCD and triangle QBC are on the same base BC, therefore

$ar\left(QBC\right)=ar\left(ABCD\right)/\mathrm{2.........}\left(ii\right)$

now from (i) and (ii), we get

$ar\left(PAB\right)=ar\left(QBC\right)$