Question

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.

Answer 1

Given: ABCD is a quadrilateral in which AC and BD are diagonals which intersect at O, such that ar (∆AOD) = ar (∆BOC)

To prove: ABCD is a trapezium.

Proof: We are given that

ar (∆AOD) = ar (∆BOC) …(i)

Adding ar (∆AOB), we get on both sides

ar (∆AOD) + ar (∆AOB) = ar (∆BOC)+ ar (∆AOB)

⇒ ar (∆ABD) = ar (∆ABC)

But triangles ABD and ABC are on the same base AB and have equal in areas.

Therefore they must lie in between the same parallels, 

i.e. AB || DC

Hence, ABCD is a trapezium.