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.

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 $\triangle DGI_2$, $\triangle DFI_1$, $\triangle I_1DI_2$ we get: $$DI_1= r_1 \sqrt{2},$$ $$DI_2= r_2 \sqrt{2},$$ $$I_1I_2= \sqrt{2(r_1^2+r_2^2)} \quad (1).$$

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: $$r_3=\sqrt{r_1^2+r_2^2} \quad (2)$$

Now let's calculate $r$, the inscribed circle radius of $\triangle ABC$, and compare with $r_3$.

See the picture below:

We know that $\triangle ADC$, $\triangle BDA$ and $\triangle BAC$ are similar, then $$\frac{r_1}{c}=\frac{r_2}{b}=\frac{r}{a}=k \quad (3)$$ So using relation $(3)$ and Pytagoras' Theorem again ($\triangle ABC$) we get: $$a^2=b^2+c^2 \Rightarrow (\frac{r}{k})^2=(\frac{r_1}{k})^2 + (\frac{r_2}{k})^2 \Rightarrow$$ $$\Rightarrow r^2=r_1^2+r_2^2 \Rightarrow r=\sqrt{r_1^2+r_2^2} \quad (4)$$

Therefore comparing $(2)$ and $(4)$ we can conclude finally that $$r=r_3.$$