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A B C D is a parallelogram whose diagona...

`A B C D` is a parallelogram whose diagonals `A C` and `B D` intersect at `Odot` A Line through `O` intersects `A B` at `P` and `D C` at `Qdot` Prove that `ar(trianglePOA)=ar(triangle QOC)dot`

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Given:
`ABCD` is a parallelogram
To prove:
`ar(trianglePOA)`= `ar(triangleGOC)`
Proof:
In` /_\ POA `and` /_\ QOC`
We have
`=>/_AOP=/_COQ ` [vertically opposite angle]
`=>AO=OC` [ Diagonals of parallalogram bisect each other]
`=>/_PAC =/_OCA` [Alternate interior angles]
`therefore/_\ POA ~= /_\QOC` [ By ASA congruece]
We know that congruent figures are equal in areas
`therefore ar ( /_\POA)=ar ( /_\QOC)`
Hence Proved.
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