In the triangle ABC, what's the measure $\angle ACB$?.
For reference: (exact copy of the question) The $\angle B$ of a triangle $ABC$ measures $\angle 16$.$"I"$ is incenter and $"E"$ is excenter relative to $BC$. If $IE = 2 AC$ , calculate $\angle ACB$.
My progress: Tried to use similarity of triangles as in another similar question but it still lacks some relationship. Here is the drawing with the relationships I found
$\angle C = 164^o - \angle A\\ BICE~ is~ cyclic\\ \triangle ACE \sim \triangle AIB \implies \frac{x}{AI}=\frac{CE}{BI}=\frac{AE}{AB}\\ \triangle AIC \sim \triangle ABE \implies \frac{x}{2x+AI}=\frac{CI}{EB}=\frac{AI}{AB}$
$\endgroup$ 41 Answer
$\begingroup$Observe,$$\angle ICE=\angle IBE=90^{\circ}\implies BECI \;\text{is cyclic}. $$Let $M$ be the midpoint of $IE$ and the center of the circumcircle of quadrilateral $BECI$. We have,$$ IM=EM=CM=AC.$$Hence,$$\angle CAM=\angle CMI=2\angle CBI=16^{\circ}$$Therefore, $$\angle ACB=180^{\circ}-\angle A-\angle B= 180^{\circ}-32^{\circ}-16^{\circ}=\boxed{132^{\circ}}$$
$\endgroup$ 4
