Question

ABCD is a parallelogram whose diagonals AC and BD intersect at O.A line through O intersects AB at P and DC at Q. Prove that $ar\left(△POA\right)=ar\left(△QOC\right).$

Answer 1

Given : In ||gm ABCD, diagonals Ac and BD intersect at O
A line through O intersect at Q

To prove : $ar\left(△POA\right)=ar\left(△QOC\right)$
Proof : In $△POA$ and $△QOC,$

$\angle AOD=\angle COQ$ (Vertically opposite angle)

OA = OC (O is mid point of AC)
$\angle APO=\angle COQ$ (Alternate interior angle)
$\therefore \left(△POA\right)\cong \left(△QOC\right)\left(byASAcriteria\right)\therefore ar\left(△POA\right)=ar\left(△QOC\right)$