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


ΔDAC and ΔDBC lie on the same base DC and 

between the same parallels AB and CD.

Ar(ΔDAC) = ar(ΔDBC)

=> ar(ΔDAC) – ar(ΔDOC) = ar(ΔDBC) – ar(ΔDOC)

=> ar(ΔAOD) = ar(ΔBOC)

Hence Proved

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