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 $\text{ar}(\triangle\text{POA})=\text{ar}(\triangle\text{QOC}).$

Answer

In triangles POA and QOC, we have$\angle\text{AOP}=\angle\text{COQ}$
$\text{AO}=\text{OC}$
$\angle\text{PAC}=\angle\text{QCA}$
So, by ASA congruence criterion, we have$\triangle\text{POA}\cong\triangle\text{QOC}$
$\Rightarrow\text{ar}(\triangle\text{POA})=\text{ar}(\triangle\text{QOC}).$

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