a geometry problem about inscribed and circumscribed circle radius. Let $\triangle ABC$ be a rectangular triangle with $m(\angle A)=90^{\circ}$ and $AD \perp BC, D\in[BC]$. Denote with $I_{1}$ the center of the circle inscribed in triangle $\triangle ADB$ and with $I_{2}$ center of the circle inscribed in triangle $\triangle ADC$. Prove that the circumscribed circle radius of the $\triangle AI_{1}I_{2}$ is equal with the inscribed circle radius of the the triangle $\triangle ABC$. 
No idea for this problem, I make a draw, and I can't do anything more.
Thanks for your help :-)
 A: See the figure below:

Let $I_1F = r_1$, $I_2G = r_2$, and $I_1E = r_3$.
Note that $GD =r_2$, $DF = r_1$, and $I_2E = r_3$.
Note also that $\triangle I_1DI_2$ is a right triangle.
So using Pytagoras' Theorem in triangles $\triangle DGI_2$, $\triangle DFI_1$, and $\triangle I_1DI_2$ we get:
$$DI_1= r_1 \sqrt{2},$$
$$DI_2= r_2 \sqrt{2},$$
$$I_1I_2= \sqrt{2(r_1^2+r_2^2)} \tag1$$
Let $m(\angle I_1AI_2)= \alpha$, as $AI_2$ is bisector of $\angle CAD$ and $AI_1$ is bisector of $\angle BAD$, we can conclude that
$$\alpha = 45 ^{\circ}$$
But if $\alpha = 45 ^{\circ}$ then $\angle I_1EI_2$ is a right angle (central angle), and using Pytagoras' Theorem in $\triangle I_1EI_2$ and equation $(1)$ we get:
$$r_3=\sqrt{r_1^2+r_2^2} \tag2$$
Now let's calculate $r$, the inscribed circle radius of $\triangle ABC$, and compare it with $r_3$.
See the picture below:

We know that $\triangle ADC$, $\triangle BDA$, and $\triangle BAC$ are similar, then
$$\frac{r_1}{c}=\frac{r_2}{b}=\frac{r}{a}=k \tag3$$
So using relation $(3)$ and Pytagoras' Theorem again ($\triangle ABC$) we get:
$$a^2=b^2+c^2 \Rightarrow \left(\frac{r}{k}\right)^2=\left(\frac{r_1}{k}\right)^2 + \left(\frac{r_2}{k}\right)^2 \Rightarrow $$
$$\Rightarrow r^2=r_1^2+r_2^2 \Rightarrow r=\sqrt{r_1^2+r_2^2}  \tag4$$
Therefore comparing $(2)$ and $(4)$ we can conclude finally that
$$r=r_3.$$
