Question

P and Q are any two points lying on the sides DC and AD respectively of a square ABCD.
Show that ar(APB) = ar(BQC).

Answer 1

$△APB\text{and parallelogram(square) ABCD lie on the same base AB and in between same parallel AB and DC.}$
$ar△APB)=\frac{1}{2}ar\text{(square}ABCD)—\left(i\right)$
Similarly,
$ar\left(△BQC\right)=\frac{1}{2}ar\text{(square ABCD) — (ii)}$
From (i) and (ii), we have
$ar\left(△APB\right)=ar\left(△BQC\right)$