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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.

Q: 15    Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that \small ar(AOD)=ar(BOC). Prove that ABCD is a trapezium.

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M manish


We have, 
ABCD is a quadrilateral and diagonals AC and BD intersect at O such that ar(\DeltaAOD) = ar (\DeltaBOC) ...........(i)

Now, add ar (\DeltaBOA) on both sides, we get

ar(\DeltaAOD) + ar (\DeltaBOA)  = ar (\DeltaBOA) + ar (\DeltaBOC)
ar (\DeltaABD) = ar (\DeltaABC)
Since the \DeltaABC and \DeltaABD lie on the same base AB and have an equal area.
Therefore, AB || CD

Hence ABCD is a trapezium.

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