O is the any point on diagonal PR of parallelogram PQRS. Prove that Ar(PSO) = Ar(PQO)

Explanation

Draw diagonal BD. Let AC and BD intersect at O. Then, O is the midpoint of BD.
`ar(triangleDPO)=ar(triangleBPO)" "[therefore " PO is a median of " triangle BPD]`.
`ar(triangleADO)=ar(triangleABO)" "[therefore " AO is a median of " triangleABD].`
`therefore ar(triangleDPO)+ar(triangleADO)=ar(triangleBPO)+ar(triangle ABO)`
`rArr ar(triangleADP)=ar(triangleABP)`.