Diagonals `A C` and `B D` of a quadrilateral `A B C D` intersect at `O` in such a way that `a r( A O D)=a r( B O C)` . Prove that `A B C D` is a trapezium.

Explanation

Given: `ABCD` is a quadrilateral where diagonals `AC` and `BD` intersect at `O`.
`⇒ar(△AOD)=ar(△BOC)`
Adding `ar(triangleODC)` on both sides,
`⇒ar(triangleAOD)+ar(triangleODC)=ar(triangleBOC)+ar(triangleODC)`
`⇒ar(triangleADC)=ar(triangleBDC)`
Now, `△ADC` and `△BDC` lie on the same base `DC` and are equal in area and they lie between the lines `AB` and `DC`
`⇒AB` parallel `DC` (Two triangles having the same base and equal areas lie between the same parallels)
In `ABCD`,
`⇒AB` parallel to `DC`
So, one pair of opposite sides is parallel,
`∴ABCD` is trapezium.
Hence Proved.