`A B C D` is a parallelogram whose `A C\ a n d\ B D` intersect at `Odot` A line through O intersects `A B\ a t\ P\ a n d\ D C\ ` at `Qdot` Prove that `a r\ (\ P O A)=\ a r\ (\ Q O C)dot`

Explanation

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.