The diagonals of a parallelogram `A B C D` intersect at a point `Odot` Through `O` , a time is drawn to intersect `A D` at `P` and `B C` at `Q` . Show that `P Q` divides the parallelogram into two parts of equal area.

Explanation

Given: The diagonals of a parallelogram `ABCD` intersect at a point `O`. Through `O`, a line is drawn to intersect `AD` at `P` and `BC` at `Q`.
To Prove: `ar(PDCQ)=ar(PQBA)`
Proof: `AC` is a diagonal of `||^(gm) ABCD`
` therefore ar(triangleABC)=ar(triangleACD)`
`=1/2ar(||^(gm) ABCD)` ....(i)
In `triangleAOP` and `triangleCOQ`,
`AO=CO` [Diagonals of a parallelogram bisect each other]
`angleAOP=angleCOQ` [Vertically opposite angles]
`angleOAP=angleOCQ` [ Alternate interior angles]
`therefore triangleAOP=triangleCOQ`
`therefore ar(triangleAOP)=ar(triangleCOQ)` [By `ASA` Congruency]
`=>ar(triangleAOP)+ar(OPDC)`
`=ar(triangleCOQ)+ar(OPDC)`
`=>1/2 ar(||^(gm) ABCD)=ar(PDCQ)`
From equation (1):
`=>ar(PQBA)=ar(PDCQ)`
`=>ar(PDCQ)=ar(PQBA)`
Hence Proved.