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

Triangles ABC and ABD are on the same base AB and between the same parallels AB and DC. Therefore, ar(ABD) = ar(ABC)
Subtract area of ΔAOB from both the sides.
ar(ABD) - ar(AOB) = ar(ABC) - ar(AOB)
⇒ ar(AOD) = ar(BOC)