| In triangle ABC, the bisector AK is drawn. It is known that the centers of the circle inscribed in triangle $ABK$ and the circle circumscribed around triangle $ABC$ coincide. Find the angles of triangle $ABC$.
Let $F$ be the point of intersection of ray $A K$ with the circumcircle of triangle $A B C$. Prove that triangle $FAB$ is isosceles.

## Solution

Let $O$ be the center of the circle, and $F$ be the point of intersection of ray $A K$ with the circumcircle. Since rays $A F$ and $A B$ form the same angles with the diameter passing through point $A$, triangle $F A B$ is isosceles.

Let $\angle B A O=\alpha$. Then

$$
\begin{gathered}
\angle O A F=\alpha, \angle B A C=4 \alpha, \angle A C B=\angle A F B=90^{\circ}-\alpha \\
\angle A K C=180^{\circ}-2 \alpha-\left(90^{\circ}-\alpha\right)=90^{\circ}-\alpha \\
\angle A B C=2 \angle A B O=2 \angle B A O=2 \alpha
\end{gathered}
$$

Therefore,

$$
90^{\circ}-\alpha=\angle A K C=\angle A B C+\angle B A K=4 \alpha
$$

Thus, $\alpha=18^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-64.jpg?height=491&width=504&top_left_y=1135&top_left_x=777)

## Answer

$72^{\circ}, 72^{\circ}, 36^{\circ}$.