Find the radius of the inscribed circle of a right triangle. Find the radius of the inscribed circle of a right triangle. The triangle's height is $\sqrt{6} + \sqrt{2}$ while the bisector of the right angle is 4.
This seemed like a generic similar triangles problem but it isn't that simple when I tried to solve it. Any help?
EDIT: Here is a picture of the problem


Find the radius $r$ of the inscribed circle of a right triangle
$ABC$ 
given its height $|CD|=h_c=\sqrt6+\sqrt2$
and bisector $|CE|=4$.
Let $I$ be the center of the inscribed circle
and $A_t,B_t,C_t$ its touching points
\begin{align} 
\triangle CED,\ \triangle IEC_t,\triangle CB_tI:\quad
\frac{h_c}{\beta_c}
&=
\frac{r}{|IE|}
=
\frac{r}{\beta_c-|CI|}
=
\frac{r}{\beta_c-r\,\sqrt2}
,\\
r&=\frac{\beta_c\,h_c}{\beta_c+h_c\,\sqrt2}
=\frac{2\,\sqrt6}3
.
\end{align}
Edit
Additionally, the side lengths of the triangle can be found as follows.
We know that
\begin{align}
2r&=a+b-c
,\\
r\,(a+b+c)&=c\,h_c
,
\end{align}
so the hypotenuse is found as
\begin{align} 
c&=
\frac{2\,h_c\,\beta_c^2}{2\,h_c^2-\beta_c^2}
=
\tfrac{4\sqrt6}3\,(1+\sqrt3)
\approx 8.92284
.
\end{align}
The sizes of the legs can be found 
from system
\begin{align}
ab&=c\,h_c
,\\
a+b&=2\,r+c
.
\end{align}
Since $a<b$, we have
\begin{align}
a&=
r+\tfrac12\,\Big(c-\sqrt{(2r+c)^2-4\,c\,h_c}\Big)
= 
\tfrac{2\sqrt6}3\,(1+\sqrt3)
,\\
b&=
r+\tfrac12\,\Big(c+\sqrt{(2r+c)^2-4\,c\,h_c}\Big)
=
2\,\sqrt2\,(1+\sqrt3)
.
\end{align}
It follows that 
$\angle CAB=30^\circ$,
$\angle ABC=60^\circ$.
