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

(N/A) We have a trapezium $ABCD$ with $AB || DC$. The diagonals $AC$ and $BD$ intersect at $O$.
Since triangles on the same base and between the same parallels have equal areas:
$\because \Delta ABD$ and $\Delta ABC$ are on the same base $AB$ and between the same parallels $AB$ and $DC$,
$\therefore \operatorname{ar}(\Delta ABD) = \operatorname{ar}(\Delta ABC)$.
Subtracting $\operatorname{ar}(\Delta AOB)$ from both sides,we get:
$\operatorname{ar}(\Delta ABD) - \operatorname{ar}(\Delta AOB) = \operatorname{ar}(\Delta ABC) - \operatorname{ar}(\Delta AOB)$
$\Rightarrow \operatorname{ar}(\Delta AOD) = \operatorname{ar}(\Delta BOC)$.