$ABCD$ 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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Updated on: 10-Oct-2022

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