euclidean-geometryplane-geometry
For reference: (Exact copy of the problem)
In the right triangle, right angle in $B$; $I$ is incenter and $E$ is excenter
for $BC$.If $AI = IE$, calculate $\angle ACB$ (answer:$53^o$)
My progress:
Drawing according to the statement and placing the relationships we arrive at…
(The dots in blue are just to make it easier to see the size of the sides)
Let $BH$ intersect $MI$ and $MN$ at $P$ and $Q$ respectively.
In two similar triangles, ratio of corresponding sides=ratio of inradii
$\small{\triangle ABH\sim\triangle BCH}\implies\frac{AB}{BC}=\frac{GH}{HF}=\frac{MH}{HN}\implies\small{\triangle ABC\sim\triangle MHN}$
Therefore, $\angle HMN=90^\circ-\theta$
$\angle MQP=45^\circ+(90^\circ-\theta)=135^\circ-\theta$
$\angle ABH=\theta, \angle BAD=45^\circ-\frac{\theta}{2}\implies\angle BPM=135^\circ-\frac{\theta}{2}$
Therefore from $\triangle PQM$, $\angle IMN=(135^\circ-\frac{\theta}{2})-(135^\circ-\theta)=\frac{\theta}{2}$
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, $$\angle ACB=180^{\circ}-\angle A-\angle B= 180^{\circ}-32^{\circ}-16^{\circ}=\boxed{132^{\circ}}$$
Best Answer
From $I$ and $E$ drop perpendiculars to $AC$. These perpendiculars will be inradius $r$ and exradius, $r_a$ respectively.
Since $AI=AE/2$, $r=r_a/2$ using similar triangles.
Using formulas for $r=\Delta / s$ and $r_a = \Delta / (s-a)$, we can obtain
$$b+c=3a$$
Combining this with $b^2-c^2=a^2$, we get $$\frac{c}{b}=\frac{4}{5} \Rightarrow \angle C = \sin^{-1}\frac{4}{5} \approx 53^\circ$$