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Diagonals A C\ a n d\ B D of a quadri...

Diagonals `A C\ a n d\ 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)dot` Prove that `A B C D` is a trapezium.

Text Solution

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Given: `ar(ΔAOD)=ar(ΔBOC)`
To 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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