3. in triangle $A B C$, let $D$ be the intersection of $B C$ with the bisector of $\triangle B A C$. The center of the circumcircle of $\triangle A B C$ coincides with the center of the inscribed circle of $\triangle A D C$. Find the angles of $\triangle A B C$.

## Solution:
As usual, let $\alpha=\angle B A C, \beta=\angle C B A$ and $\gamma=\angle A C B$. The common center of the two circles mentioned is $O$. The lines from $O$ to $A, B$ and $C$ are drawn as auxiliary lines.

They now follow one after the other (reasons in brackets):

(1) $\angle A C O=\angle O C B=\frac{\gamma}{2}(O$ is the center of the inscribed circle of $\triangle A D C)$

(2) $\angle O A C=\frac{\alpha}{4}$ (same reason, moreover $A D$ is angle bisector of $\angle B A C$ )

(3) The triangles $O A B, O B C$ and $O C A$ are isosceles. ( $O$ is the center of the circumcircle of $\triangle A B C$, therefore $O A=O B=O C$).

(4) $\angle C B O=\frac{\gamma}{2}$ (since $\triangle O B C$ is isosceles).

(5) $\angle O B A=\frac{3 \alpha}{4}$ (since $\triangle O A B$ is isosceles).

(6) $\frac{\alpha}{4}=\frac{\gamma}{2}$ (since $\triangle O C A$ is isosceles).

This enables us to express $\gamma=\frac{\alpha}{2}$ as well as $\beta$ by $\alpha$:

$$
\beta=\angle C B O+\angle O B A=\frac{\gamma}{2}+\frac{3 \alpha}{4}=\alpha
$$

Because of the sum of angles in $\triangle A B C$, $\alpha$ is thus uniquely determined:

$$
180^{\circ}=\alpha+\beta+\gamma=\alpha+\alpha+\frac{\alpha}{2}=\frac{5}{2} \alpha
$$

Therefore, $\alpha=\beta=72^{\circ}$ and $\gamma=36^{\circ}$ must apply.

The triangle is shown in the illustration.

![](https://cdn.mathpix.com/cropped/2024_04_28_355f5ae1f764b2e3feacg-3.jpg?height=719&width=673&top_left_y=1348&top_left_x=726)