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Diagonals A C and B D of a quadrilateral...

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.

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Given: `ar(/_\AOD)=ar(/_\BOC)`
To prove: `ABCD` is a trapezium.
Proof: `ar(/_\AOD)=ar(/_\BOC)`
`=>ar(/_\AOD)+ar(/_\AOB)=ar(/_\BOC)+ar(/_\AOB)`
`=>ar(/_\ADB)=ar(/_\ACB)`
Areas of `/_\ADB` and `/_\ACB` 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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