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Diagonals `A C and B D` of a trapezium `A B C D` with `A B||D C` intersect each other at `O` . Prove that `ar ( AOD)= ar ( BOC)` .

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Given `ABCD` is a trapezium the diagonal `AC` and `BD` with `AB∥CD` intersect each other at O.
`AB|\ |CD`
Then `/_\ADB`and `/_\ACB` lie on same base `AB` and between two parallel line `AB` and `CD`
`therefore area/_\ADB=area/_\ACB`
Subtract the area of triangle `AOB` on both sides we get
`⇒area/_\ADB−area/_\AOB=area/_\ACB−area/_\AOB`
As per figure
`⇒area(/_\ AOD)=area(/_\ BOC)`
Hence proved.
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