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

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Solution

It can be observed that ΔDAC and ΔDBC lie on the same base DC and between the same parallels AB and CD.

∴ Area (ΔDAC) = Area (ΔDBC)

⇒ Area (ΔDAC) − Area (ΔDOC) = Area (ΔDBC) − Area (ΔDOC)

⇒ Area (ΔAOD) = Area (ΔBOC)

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Diagonals AC and BD of a trapezium ABCD with AB $∥$ DC intersect each other at O.
Prove that ar ( $△$ AOD) = ar ( $△$ BOC). [1 MARK]

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