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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).

It can be observed that $$\triangle{DAC}$$ and $$\triangle{DBC}$$ lie on the same base DC and between the same parallels AB and CD.
$$\therefore$$ Area $$(\triangle{DAC}$$) = Area ($$\triangle{DBC}$$)
$$\Rightarrow Area (\triangle{DAC}) - Area (\triangle{DOC}) = Area (\triangle{DBC}) - Area (\triangle{DOC})$$
Area ($$\triangle{AOD}$$) = Area ($$\triangle{BOC}$$)