What's the measure of the $\angle ACB$?

Question

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)

Answer 1

$ \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$

$\therefore \small \angle ACB = 53^\circ$

Answer 2

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 $$\frac{c}{b}=\frac{4}{5} \Rightarrow \angle C = \sin^{-1}\frac{4}{5} \approx 53^\circ$$