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

It can be observed that \(\triangle{DAC}\) and \(\triangle{DBC}\) lie on the same base DC and between the same parallels AB and CD.

\(\therefore \) Area \( (\triangle{DAC}\)) = Area (\(\triangle{DBC}\))
\(\Rightarrow Area (\triangle{DAC}) - Area (\triangle{DOC}) = Area (\triangle{DBC}) - Area (\triangle{DOC})\)
Area (\(\triangle{AOD}\)) = Area (\(\triangle{BOC}\))
Hence, proved.