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.
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Best Answer
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 $$r=r_3.$$