Continuing the ray $B C$ to intersect the circumcircle of triangle $B B_{1} C_{1}$, we obtain point $K$ (see the figure). The inscribed angles $\angle C_{1} B B_{1}$ and $\angle K B B_{1}$ are equal (since $B B_{1}$ is the angle bisector), which means the arcs they subtend, $B_{1} C_{1}$ and $B_{1} K$, are equal. Points $K$ and $C_{1}$ lie on the circumcircle (described around triangle $B B_{1} C_{1}$), whose center lies on the line $A C$. Therefore, $K$ and $C_{1}$ are symmetric with respect to the line $A C$. We obtain the equality of three angles: $\angle B C C_{1} = \angle C_{1} C B_{1} = \angle B_{1} C K$. The sum of these angles is $180^{\circ}$, so each of them is $60^{\circ}$, and $\angle A C B = \angle B C C_{1} + \angle C_{1} C B_{1} = 120^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1b17058527caf5b5d058g-39.jpg?height=366&width=643&top_left_y=2395&top_left_x=702)

Comments. 1. It is easy to show that the center $O$ of the circle can only lie on the extension of segment $A C$ beyond point $C$, and thus the line $B C$ intersects the circle exactly as shown in the figure.

2. Solutions based on the fact that points $B, C, O, C_{1}$ lie on the same circle, or that the circumcircle of triangle $B C_{1} B_{1}$ is the Apollonian circle for points $A$ and $C$, are also possible.

## Answer

$120^{\circ}$.
Zaslavsky A.A.

In triangle $A B C$, the bisectors $B B_{1}$ and $C C_{1}$ are drawn. It is known that the center of the circumcircle of triangle $B B_{1} C_{1}$ lies on the line $A C$. Find the angle $C$ of the triangle.