Proving angle equality in a circle

Question

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.

Answer 1

Hint:

• $|\angle AOB| = 2 |\angle ACB|$, $|\angle COD| = 2|\angle CAD|$,
• $|\angle AEC| + |\angle ECA| + |\angle CAE| = \pi$.

I hope this helps $\ddot\smile$

Answer 2

By the theorem about the angle of two crossing cords in a circle , we have that

$$\angle AEB=\frac12\left(\widehat{AB}+\widehat{CD}\right)$$

On the other hand, by definition of central angle:

$$\angle AOB=\widehat{AB}\;,\;\;\;\angle COD=\widehat{CD}$$

Well, there you are! And note that the inscribed angles theorem plays no role win this demonstration.