Question

A point O is taken in the interior of a parallelogram ABCD. Prove that the sum of the areas of the triangles OAB and OCD is constant at any choice of the point O.

Answer 1

Area of the parallelogram is $|AB|.(p+q)$

Area of $△OAB=\frac{1}{2}|AB|.p$

Area of $△OCD=\frac{1}{2}|CD|.q=\frac{1}{2}|AB|.q$

Thus sum of areas of $△OAB$ and $△OCD=\frac{1}{2}|AB|.(p+q)=\frac{1}{2}\times$ the area of the parallelogram which is a constant