<s>[INST] 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 $\angle IEP$

 [/INST] 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, $$\small{\frac {5}{\sin 26.5^\circ}=\frac{2}{\sin \theta}}$$ $$\small{\implies \theta =\arcsin \left(\frac {2\sqrt 5}{25}\right)\approx 10.30^\circ=\angle IEP}$$

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