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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