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
$\angle C = 164^o – \angle A\\
BICE~ is~ cyclic\\
\triangle ACE \sim \triangle AIB \implies \frac{x}{AI}=\frac{CE}{BI}=\frac{AE}{AB}\\
\triangle AIC \sim \triangle ABE \implies \frac{x}{2x+AI}=\frac{CI}{EB}=\frac{AI}{AB}$
Best Answer
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, $$\angle ACB=180^{\circ}-\angle A-\angle B= 180^{\circ}-32^{\circ}-16^{\circ}=\boxed{132^{\circ}}$$