geometry
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……
Build a parallelogram $ABDE$. Then $$\angle DEA = \angle ABD = 135^\circ = \angle DCA,$$ hence $ADEC$ is cyclic. Moreover, $\angle EAD = \angle BDA = x$, hence $$\angle CAE = \angle CAD - \angle EAD = 2x-x=x.$$ As a consequence, $CE=DE$ because the angles subtended by arcs $CE$, $DE$ of the red circle are equal.
Now, it is easy to calculate that angle $CBD$ equals $90^\circ + x$ and that the concave angle $CED$ equals $180^\circ + 2x$. Since $CE=DE$, it follows that $B$ lies on the circle with center $E$ and radius $CE=DE$. Hence $\angle DEB = 2\angle DCB = 90^\circ$. Since $DE=BE$, the right triangle $DEB$ is isosceles.
Let $F$ be the midpoint of $BE$. Let $G$ be the midpoint of $BD$ and $H$ be the midpoint of $BG$. Easy to see that triangles $GFB$ and $GFH$ are isosceles right triangles, hence $HF = HB = \frac 12BG = \frac 14 BD$, so $HD = BD - BH = 4HF - HF = 3HF$. This yields $\tan x = \frac{HF}{HD} = \frac 13$ so the answer is $$x = \arctan \frac 13.$$
Given $\triangle ABC$ is right triangle, you can easily show that
$BC = R + r$, where $R$ and $r$ are respectively the radii of the excircle on side $BC$ and the incircle.
So, $CP = R$.
Now $\triangle AEH$ is a special triangle with ratio of sides $EH:AH = 1:2$
$\implies R: (R + r + AP) = 1: 2$
$R = r + AP = AB = EH$
Also, $BC = PH = R + r$
So $\triangle ABC \sim \triangle EHP$
$\angle EPH = 37^\circ$
$\therefore \angle IEP = 37^\circ - \frac{53^\circ}{2} = 10.5^\circ$
Best Answer
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$