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

Triangles ABD and ABC are on the same base AB and between same parallels, AB∥ CD.
Hence, ar (ABD) = ar(ABC)
[∵ Triangles on the same base (or equal bases) and between the same parallels are equal in area.]
Subtracting ar (ABO) from both of sides
ar(ABD) – ar(ABO) = ar(ABC) – ar(ABO)
⇒ ar(AOD) = ar(BOC)