In the adjoining figure, PQRS and PABC are two parallelogram of equal area. Prove that QC||BR.

Explanation

Let BC and QR itersect at O.
Join BQ, QC, CR and RB.
Now, `ar("||gmPQRS")=ar("||gmPABC") " " ("given)`
`rArr ar("||gm PQRS")-ar("||gm QOCP")`
`=ar("||gm PABC)-ar("||gm QOCP")`
`rArr ar("||gm ORSC")=ar("||gm ABOQ")`
`rArr 2xxar(triangleORC) =2xxar(triangleOBQ)rArr(triangleORC)=ar(triangleOBQ)`
`rArr ar(triangleORC)+ar(triangleOBR)=ar(triangleOBQ)+ar(triangleOBR)`
`rArr ar(triangleBRC)=ar(BRQ)`.
Now, `triangleBRQ` being on the same base BR and equal in area, they must be between the same parallels.
`therefore QC||BR.`