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

(N/A) We have a quadrilateral $ABCD$ and its diagonals $AC$ and $BD$ intersect at $O$ such that $\operatorname{ar}(\Delta AOD) = \operatorname{ar}(\Delta BOC)$.
Now,$\operatorname{ar}(\Delta AOD) = \operatorname{ar}(\Delta BOC)$ [Given].
Adding $\operatorname{ar}(\Delta AOB)$ to both sides,we have:
$\operatorname{ar}(\Delta AOD) + \operatorname{ar}(\Delta AOB) = \operatorname{ar}(\Delta BOC) + \operatorname{ar}(\Delta AOB)$
$\Rightarrow \operatorname{ar}(\Delta ABD) = \operatorname{ar}(\Delta ABC)$.
Since these two triangles $\Delta ABD$ and $\Delta ABC$ are on the same base $AB$ and have equal areas,they must lie between the same parallel lines.
Therefore,$AB \parallel DC$.
Since $ABCD$ is a quadrilateral having a pair of opposite sides parallel,$ABCD$ is a trapezium.