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# 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.

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.

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