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.

Solution

Given:

ar(ΔAOD) = ar(ΔBOC)

Prove:

ABCD is a trapezium.

Proof:

ar(ΔAOD) = ar(ΔBOC)

=> ar(ΔAOD) + ar(ΔAOB) = ar(ΔBOC)+ar(ΔAOB)

=> ar(ΔADB) = ar(ΔACB)

Areas of ΔADB and ΔACB are equal.

Thus

They should be lying between the same parallel lines.

Hence AB ∥ CD

Therefore, ABCD is a trapezium.

Hence Proved

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