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In the given figure, $O$ is the center of the circle and $$ \angle AOB =120$$

enter image description here

How could I find the measure of $\angle AEB$?

Thanks in advance.

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With $A,B,O$ fixed, $C$ and $D$ determine $E$, so $\angle AEB$ depends on $C$ and $D$, no? Then what are those? – anon Apr 22 '12 at 3:34
You cannot find <AEB. The most we can know is that <ADB = <ACB = <AOB/2. Imagine holding point C still while varying point D. The angle <AEB changes. – user29660 Apr 22 '12 at 3:43
Agree with anon, more info needed. Angle ABD is 60 degrees, so angle AEB is 60 plus angle EAD, which angle depends on where C and D are. – Gerry Myerson Apr 22 '12 at 3:43
It seems we are missing a piece of information, such as the $\angle$ for $C$ or $D.$ Also, what have you tried so far? – Joe Apr 22 '12 at 3:44
To be more specific about the ambiguity: if $AC$ and $BD$ were diameters (containing $O$), then $\angle AEB = \angle AOB = 120^{\circ}$. On the other hand, if $C$ and $D$ were to coincide (along the longer arc $AB$), then $E$ would coincide with them, so that $\angle AEB = \angle ACB = \angle ADB = 60^{\circ}$ (by the Inscribed Angle Theorem ). With $C$ and $D$ elsewhere, $\angle AEB$ is something else. This, even though the $\angle C$ and $\angle D$ --with $C$ and $D$ along the longer arc $AB$-- are always $60^{\circ}$ (by the IAT). – Blue Apr 22 '12 at 6:35
up vote 3 down vote accepted

As discussed in the comments, the measure of the angle depends on the positions of $C$ and $D$:


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