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 $$\triangle DGI_2$$, $$\triangle DFI_1$$, 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 $$r=r_3.$$