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# 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.

Given:
Area ($$\triangle{AOD}$$) = Area ($$\triangle{BOC}$$)

$$\Rightarrow$$ Area ($$\triangle{AOD}$$) + Area ($$\triangle{AOB}$$) = ($$\triangle{BOC}$$) + Area ($$\triangle{AOB}$$)

$$\therefore$$ Area ($$\triangle{ADB}$$) = Area ($$\triangle{ACB}$$)

We know that triangles on the same base having area equal to each other lie between the same parallels.

$$\therefore$$ these triangles, $$\triangle{ADB}$$ and $$\triangle{ACB}$$, are lying between the same parallels.i.e., AB || CD

$$\therefore$$ ABCD is a trapezium.
Hence, proved.