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Answer :
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.