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$ABCD$ 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 $ar(\triangle POA) = ar(\triangle QOC)$.
Given:
$ABCD$ is a parallelogram whose diagonals $AC$ and $BD$ intersect at $O$. A line through $O$ intersects $AB$ at $P$ and $DC$ at $Q$.
To do:
We have to prove that $ar(\triangle POA) = ar(\triangle QOC)$.
Solution:
In $\triangle POA$ and $\triangle QOC$,
$OA = OC$ ($O$ is the mid-point of $AC$)
$\angle AOD = \angle COQ$ (Vertically opposite angles)
$\angle APO = \angle CQO$ (Alternate angles)
Therefore, by AAS axiom,
$ar(\triangle POA) \cong ar(\triangle QOC)$
This implies,
$ar(\triangle POA) = ar(\triangle QOC)$
Hence proved.
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