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)
2 Answers
$\begingroup$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$$\frac{c}{b}=\frac{4}{5} \Rightarrow \angle C = \sin^{-1}\frac{4}{5} \approx 53^\circ$$
$\endgroup$ 0 $\begingroup$$ \small \triangle ABE \sim \triangle AIC$
So, $ \small BE:AB = CI:AI$
But $ \small CI = CE = \frac{IE}{\sqrt2} = \frac{AI}{ \sqrt2}$.
$ \implies \small BE = \frac{AB}{\sqrt2} \tag1$
Also, $ \small \triangle AIB \sim \triangle ACE$,
So, $ \small BI:AB = CE:AE$
But $ \small AE = 2 AI$ and hence $ \small CE:AE = 1: 2 \sqrt2$.
$ \small \implies BI = \frac{AB}{2 \sqrt2} \tag2$
From $(1)$ and $(2)$, $ \small BE = 2 BI $
$ \small \triangle IBE$ is a special right triangle with perp sides in ratio $1:2$.
So we have, $ \small \angle IEB = \frac{\angle C}{2} = 26.5^\circ$
$\therefore \small \angle ACB = 53^\circ$
$\endgroup$ 0
