`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

Construction: Join `AC`.
Let `AC` and `BD` intersect space at space straight `P`.
Now, `P` is the midpoint of diagonals `AC` and `BD` since diagonals of a parallelogram bisect each other.
`BP` is the median of `triangleABC`.
Area of `trianglePAB=`Area of `trianglePCB`......(1)
Similarly `OP` is the median of `triangleOAC`.
Area of `triangleOAP=`Area of `triangleOCP`...... (2)
Adding (1) and (2), we get
Area of `trianglePAB+`Area of `triangleOAP=`Area of `trianglePCB+`Area of `triangleOCP`
Area of `triangleOAB=`Area of `triangleOBC`
Hence Proved.