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.

ar(AOD) = ar(BOC) [∵ Given]
adding ar(AOB) both of sides
ar(AOD) + ar(AOB) = ar(BOC) + ar(AOB)
⇒ ar(ABD) = ar(ABC)
ΔABC and ΔABC are on the same base AB and ar(ABD) = ar(ABC).
Therefore, AB ∥ DC
[∵ Triangles on the same base (or equal bases) and having equal areas lie between

the same parallels.]
Hence, ABCD is a trapezium.