What is the value of the $\measuredangle IEP$ in the figure below? In a right triangle ABC straight at B, $\measuredangle C = 37^\circ. $If $E$ is an excenter in relation to BC, I is an incenter and $P$ is the point of tangency of the circle inscribed with AC.
calculate $\measuredangle IEP$
My progres::
I made the drawing and marked the relationships found
Let $KE \perp CB$
AE is angle bissector = $\frac{53}{2} \implies \measuredangle AEJ = \frac{137}{2}\\
JEK =90^\circ\implies \measuredangle KEI = \frac{53}{2}$
but I still haven't found the link to find $\angle IEP$

 A: HINT:
$\small{\triangle ABC}$ is a $\small{3,4,5}$ triangle and $\small{\triangle AJE}$ is a right triangle with perpendicular sides in the ratio $\small{1:2}$. Therefore say $\small{AB=3, BC=4, AC=5}$ and $\small{BJ=JE=3}$. This leads $\small{AE=3\sqrt 5}$.
Let $\small{AB}$ touches incircle at $\small Q$, then define $\small{AP=x, BQ=y, CP=z}$. Then $$\small{\begin{align}x+y &=3 \\ y+z &=4 \\ x+z&=5 \\ \end{align}}$$ By solving equations, we get $\small{x=2, y=1, z=3}$.
So we know $\small{AP=2, AE=3\sqrt 5}$ and $\small{\angle PAE=26.5^\circ}$. Thus by applying cosine rule and sine rule to $\small{\triangle APE}$ we can evaluate $\small{\angle IEP}$.

 (Reminder: $\small{\tan 26.5^\circ\approx\frac 12}$) By cosine rule, $$\small{\cos 26.5^\circ=\frac{2}{\sqrt 5}=\frac{2^2+(3\sqrt 5)^2-(PE)^2}{2\cdot 2\cdot 3\sqrt 5}}$$ $$\small{\implies PE=5}$$ By sine rule, $$\small{\frac {5}{\sin 26.5^\circ}=\frac{2}{\sin \theta}}$$ $$\small{\implies \theta =\arcsin \left(\frac {2\sqrt 5}{25}\right)\approx 10.30^\circ=\angle IEP}$$

A: 
Hint for geometric solution:
If you prove that  triangle MPC is isosceles, i.e MP=MC, then:
$\angle MPC=37^o= \frac {53}2+\angle PEI$
which gives $\angle PEI=10.5^o$
A: Given $\triangle ABC$ is right triangle, you can easily show that
$BC = R + r$, where $R$ and $r$ are respectively the radii of the excircle on side $BC$ and the incircle.
So, $CP = R$.
Now $\triangle AEH$ is a special triangle with ratio of sides $EH:AH = 1:2$
$\implies R: (R + r + AP) = 1: 2$
$R = r + AP = AB = EH$
Also, $BC = PH = R + r$
So $\triangle ABC \sim \triangle EHP$
$\angle EPH = 37^\circ$
$\therefore \angle IEP = 37^\circ - \frac{53^\circ}{2} = 10.5^\circ$
