Question

$P$ and $Q$ 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

Area of triangle between same base Ab and same parallel line are half of parallelogram

$\begin{array}{c}ar\Delta APB=\frac{1}{2}arABCD\to \left(i\right)\\ simialarlyonbaseBC\\ ar\Delta BQC=\frac{1}{2}arABCD\to \left(ii\right)\\ from\left(i\right)\&\left(ii\right)\\ ar\Delta APB=ar\Delta BQC\end{array}$