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# Diagonals $ \mathrm{AC} $ and $ \mathrm{BD} $ of a trapezium $ \mathrm{ABCD} $ with $ \mathrm{AB} \| \mathrm{DC} $ intersect each other at $ \mathrm{O} $. Prove that ar $ (\mathrm{AOD})=\operatorname{ar}(\mathrm{BOC}) $.

Given:

Diagonals \( \mathrm{AC} \) and \( \mathrm{BD} \) of a trapezium \( \mathrm{ABCD} \) with \( \mathrm{AB} \| \mathrm{DC} \) intersect each other at \( \mathrm{O} \).

To do:

We have to prove that ar \( (\mathrm{AOD})=\operatorname{ar}(\mathrm{BOC}) \).

Solution:

$\triangle ABC$ and $\triangle ABD$ lie on the same base $AB$ and between the parallels $AB$ and $CD$.

This implies,

$ar (\triangle ABD) = ar (\triangle ABC)$

Subtracting $ar (\triangle AOB)$ from both sides, we get,

$ar (\triangle ABD) - ar (\triangle AOB) = ar (\triangle ABC) - ar (\triangle AOB)$

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

Hence proved.

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