$ABCD$ 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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Updated on: 10-Oct-2022

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