# Diagonals $\mathrm{AC}$ and $\mathrm{BD}$ of a quadrilateral $A B C D$ intersect at $O$ in such a way that ar $(\mathrm{AOD})=\operatorname{ar}(\mathrm{BOC})$. Prove that $\mathrm{ABCD}$ is a trapezium.

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Given:

Diagonals $\mathrm{AC}$ and $\mathrm{BD}$ of a quadrilateral $A B C D$ intersect at $O$ in such a way that ar $(\mathrm{AOD})=\operatorname{ar}(\mathrm{BOC})$.

To do:

We have to prove that $\mathrm{ABCD}$ is a trapezium.

Solution:

$ar(\mathrm{AOD})=\operatorname{ar}(\mathrm{BOC})$

Adding $ar(\triangle AOB)$ on both sides, we get,

$ar(\triangle AOD)+ar(\triangle AOB)=ar(\triangle BOC)+ar(\triangle AOB)$

$ar(\triangle ADB)=ar(\triangle ACB)$

This implies,

$ar(\triangle ADB)$ and $ar(\triangle ACB)$ lie between the same parallel lines.

Therefore,

$AB \| DC$

This implies,

$ABCD$ is a trapezium.

Hence proved.

Updated on 10-Oct-2022 13:42:12