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

Given:
Area (\(\triangle{AOD}\)) = Area (\(\triangle{BOC}\))

\(\Rightarrow \) Area (\(\triangle{AOD}\)) + Area (\(\triangle{AOB}\)) = (\(\triangle{BOC}\)) + Area (\(\triangle{AOB}\))

\(\therefore \) Area (\(\triangle{ADB}\)) = Area (\(\triangle{ACB}\))

We know that triangles on the same base having area equal to each other lie between the same parallels.

\(\therefore \) these triangles, \(\triangle{ADB}\) and \(\triangle{ACB}\), are lying between the same parallels.i.e., AB || CD

\(\therefore \) ABCD is a trapezium.
Hence, proved.