`O` is any point on the diagonal `B D` of the parallelogram `A B C Ddot` Prove that `a r( O A B)=a r( O B C)`

Explanation

Given: `O` is any point on the diagonal `BD` of the parallelogram `ABCD`
To prove: `ar(/_\OAB)=ar(/_\OBC)`
Proof: Join `AC`, which intersects `BD` at `P`. Since the diagonals of a parallelogram bisect each other.
`therefore P` is the mid-point of both `AC` and `BD`.
Now,
In `/_\ABC`, `BP` is the median.
`:. ar(/_\ABP)=ar(/_\CBP)`.......(1)
In `/_\OAC`, `OP` is the median.
`:. ar(/_\OAP)=ar(/_\OCP)` ......(2)
On adding equation (1) and equation (2), we get:
`ar(/_\ABP)+ar(/_\OAP)=ar(/_\CBP)+ar(/_\OCP)`
`=>ar(/_\OAB)=ar(/_\OBC)`.