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 $\angle IEP$
Best Answer
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 $\triangle ABC \sim \triangle EHP$
$\angle EPH = 37^\circ$
$\therefore \angle IEP = 37^\circ - \frac{53^\circ}{2} = 10.5^\circ$