What's the measure of angle ∡IEO? For reference: In a triangle ABC, ∡B=120º.
"I" ⇒ Incenter of △ ABC.
"O"⇒ Circumcenter of △ ABC
"E" ⇒ Excenter of △ ABC relative to side BC.
Calculate the m∡IEO.
My progress:
I made the figure, marked the possible angles and drew some auxiliary lines to try to reach the solution......
 A: I refer to internal angles of $\triangle ABC$ as $\angle A, \angle B$ and $\angle C$
$\angle OAE = \angle 30^\circ + \angle A = \angle 30^\circ + \frac{1}{2} (60^\circ - \angle C)$
$ \angle OAE = 60^\circ - \frac{\angle C}{2} \tag1$
$\angle OCE = 30^0 + \angle ACE = 30^\circ +(90^\circ + \frac{\angle C}{2})$
$ \angle OCE= 120^\circ + \frac{\angle C}{2} \tag2$
So, $\angle OAE + \angle OCE = 180^\circ ~ $ and it follows that quadrilateral $OAEC$ is cyclic.
$ \therefore \angle IEO = \angle ACO = 30^\circ$
A: 
Consider the fact that the bisector of $\angle ABC$ passes I and O where it intersects the perpendicular bisector of AC. So ABC is isosceles and we have:
$\angle BAC=\angle BCA=30^o$
$\Rightarrow \angle BCF=150^o$
$\Rightarrow \angle BCE=75^o$
$\Rightarrow \angle ECA=105^o$
$\Rightarrow \angle AEC=60^o$
$\angle BEA=\angle BAE=15^o$
$\Rightarrow \angle BEC=60+15=75^o$
$\Rightarrow \angle CBE=30^o$
that means:
$\angle CBO=\angle OBC+\angle CBE=60+30=9^o$
Also $BE=BC=BO$
$\Rightarrow \angle BEO=45^o$
which gives :
$\angle IEO=45-15=30^o$
