`p\ A N D\ q` are any two points lying on the sides `D C\ a n d\ A D` respectively of a parallelogram `A B C Ddot\ ` Show that `a r( A P B)=a r\ ( B Q C)dot`

Explanation

Since `/_\ APB` and parallelogram `ABCD` are on the same base `AB` and between the same parallels.
Therefore,
`ar(/_\APB)=1/2 ar( |\ |gm ABCD) .......(1)`
Similarly,
`ar(/_\ BQC)=1/2 ar( |\ | gmABCD) ......(2)`
From `1` and `2`, we get,
`ar(/_\ APB)=ar(/_\ BQC)`
Hence proved.