In the figure, $ABCD$ is a trapezium in which $AB \parallel DC$. Prove that $ar( \triangle AOD = ar(\triangle BOC)$.
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Given:
$ABCD$ is a trapezium in which $AB \parallel DC$.
To do:
We have to prove that $ar( \triangle AOD = ar(\triangle BOC)$.
Solution:
In trapezium $ABCD$, diagonals $AC$ and $BD$ intersect each other at $O$.
$\triangle ADB$ and $\triangle ACB$ are on the same base $AB$ and between the same parallels.
Therefore,
$ar(\triangle ADB = ar(\triangle ACD)$
Subtracting $ar(\triangle AOB)$ from both sides,
$ar(\triangle ADB) - ar(\triangle AOB) = ar(\triangle ACD) - ar(\triangle AOB)$
$ar(\triangle AOD) = ar(\triangle BOC)$
Hence proved.
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