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
$$r=r_3.$$
In the diagram above $A$ is the centre of the circle and $CB$ is thus the diameter. Point $D$ is an arbitrary point in the circumference.
In $\triangle ACD$ $\angle CDA = \alpha$ since $AC = AD$
Thus
$$2 \cdot \alpha + \theta = 180 ^\circ$$
In $\triangle ADB$ $\angle ADB = \beta$ since $AB = AD$
Thus
$$2 \cdot \beta + (180^\circ - \theta) = 180 ^\circ$$
Add both these together
$$2 \cdot \alpha + 2 \cdot \beta + 180 ^\circ = 360 ^\circ$$
Which we can easily rearrange to show:
$$ \alpha + \beta = 90 ^\circ $$
We have thus proved $\angle CDB$ is a right angle as required.
Best Answer
Let be the point $O$ the center of the circle and $\overline{\rm AB}$ the diameter of the circle. To show that the $ \bigtriangleup ABC$ is a right triangle, we have to show that the $\angle BAC$ is a right triangle. Then we have that the $\angle BOC$ is a straight angle, so its measure is $180^{\circ}$.
$$m\angle BOC=2\cdot m\angle BAC$$ $$180^{\circ}=2\cdot m\angle BAC$$ $$\frac{180^{\circ}}{2}=\frac{2}{2} \cdot m\angle BAC$$ $$90^{\circ}=m\angle BAC$$
Therefore the $m\angle BAC$ is a right triangle and therefore $\bigtriangleup ABC$ is a right triangle.
Then, if we draw a segment from point $O$ to point $A$ and from point $O$ to point $C$ is formed $\bigtriangleup OAC$ and is gonna be an equilateral triangle. Because $\overline{\rm OA}$ and $\overline{\rm OC}$ are also radius of the circle then $OA=1$ and $OC=1$. Therefore, $AC=1$.
Last,because $\bigtriangleup ABC$ is a right triangle the Pythagoras Theorem is satisfied and we can use it to find $AB$. If the radius of the circle is $1$ then the diameter $\overline{\rm BC}$ measures $2$.
$$b^2+c^2=a^2$$ $$1^2+c^2=2^2$$ $$1+c^2=4$$ $$c^2=4-1$$ $$c^2=3$$ $$\sqrt{c^2}=\sqrt{3}$$ $$c=\sqrt{3}$$
Therefore, $AB=\sqrt{3}$