Let $ABDC$ be a quadrilateral inscribed in a circle with centre $O$. If the diagonals $AD$ and $BC$ intersect at point $E$, then we have to prove that sum of angles $AOB$ and $COD$ is equal to twice angle $AEB$.
It obviously has everything to do with the inscribed angle theorem. But the figure becmes very messy once I draw it. I figured that if we omit the sides of the quadrilateral ,the theorem would still remain the same. But I cannot see where to start. For now I only want a hint on where to start.Unfortunately I have no good work to show to you guys.