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$ABCD$ is a parallelogram whose diagonals intersect at $O$. If $P$ is any point on $BO$, prove that $ar(\triangle ADO) = ar(\triangle CDO)$.
Given:
$ABCD$ is a parallelogram whose diagonals intersect at $O$.
$P$ is a point on $BO$.
To do:
We have to prove that $ar(\triangle ADO) = ar(\triangle CDO)$.
Solution:
Join $AP$ and $CP$.
In $\triangle ADC$,
$O$ is the mid point of $AC$
This implies,
$ar(\triangle ADO) = ar(\triangle CDO)$
Hence proved.
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