# Proving angle equality in a circle

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.

• if you talk of a quadrilateral $\,ABCD\,$ , it is customary to refer to the vertices in that order, and thus the diagonals are $\,AC\,,\,BD\,$ and not what you wrote. Please do check this and either tell us what's the order of the vertices or correct your question – DonAntonio Aug 7 '13 at 17:14
• @DonAntonio,Order of vertices :ABDC (clockwise) – rah4927 Aug 7 '13 at 17:16
• Ok @rahul, thanks. That doesn't make any sense whatsoever in what I'm used to do, but perhaps it is so designed as to remark to students not to get used to these or those names. – DonAntonio Aug 7 '13 at 17:41
• @DonAntonio,you should be thanking me for not putting in the original question.It took me a few minutes before I could understand anything.That would be an ideal example to show to students what not to get used to. – rah4927 Aug 7 '13 at 17:52

Hint:

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

I hope this helps $\ddot\smile$

• E is the point of intersection of two diagonals.The latter point does not seem to make sense. – rah4927 Aug 7 '13 at 16:58
• @rahul I missed that you switched names of $C$ and $D$, i.e. your quadrilateral is $ABDC$, not the usual $ABCD$ (fixed). Such naming isn't wrong, but I would advise you against it, it's easy to make mistakes (e.g. as I did). – dtldarek Aug 7 '13 at 17:26
• We add the two equations on your first hint and get angles AOB+COD=2(ACB+CAD).But of course ,since CB is a straight line ,angle AEB =ACB+CAD,thus conpleting the proof.Thought I'd come back and complete the proof. – rah4927 Aug 9 '13 at 9:44

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.