9.8. A circle is circumscribed around triangle $A B C$. Let $A_{0}$ and $C_{0}$ be the midpoints of its arcs $B C$ and $A B$, not containing vertices $A$ and $C$. It turns out that the segment $A_{0} C_{0}$ is tangent to the circle inscribed in triangle $A B C$. Find the angle $B$.
9.8. Let the incircle of triangle $ABC$ be denoted by $\omega$, its center by $I$, and the points of tangency of $\omega$ with $AC$ and $A_0 C_0$ by $K$ and $L$ respectively. Then, in triangles $IKC$ and $ILA_0$, angles $IKC$ and $ILA_0$ are right angles (as angles between a radius and a tangent), and $IK = IL$, since both segments are radii of $\omega$ (see Fig. 6). Also, since points $A, I, A_0$ lie on the angle bisector of angle $A$, and points $C, I, C_0$ lie on the angle bisector of angle $C$, we have $\angle ICK = \angle C_0 CA = \angle C_0 A_0 A = \angle LA_0 I$.

![](https://cdn.mathpix.com/cropped/2024_05_06_695363564d40540a0226g-2.jpg?height=374&width=434&top_left_y=1638&top_left_x=1459)

Fig. 6 (since angles $\angle C_0 A_0 A$ and $\angle C_0 CA$ subtend the same arc of the circumcircle $C_0 A$). Therefore, right triangles $IKC$ and $ILA_0$ are congruent by a leg and an acute angle, hence $IA_0 = IC$, which means triangle $IA_0 C$ is

![](https://cdn.mathpix.com/cropped/2024_05_06_695363564d40540a0226g-2.jpg?height=86&width=1625&top_left_y=2187&top_left_x=250)
$=\angle A_0 CC_0 \Rightarrow \angle A_0 IC = \angle A_0 CI \Rightarrow A_0 I = A_0 C$. Thus, $\triangle A_0 IC$ is equilateral. Therefore, $\angle ABC = \angle AA_0 C = 60^\circ$.

Remark. If we consider an arbitrary triangle $ABC$ where $\angle B = 60^\circ$, then the segment $A_0 C_0$ is tangent to $\omega$. Indeed, by reversing all the arguments (taking $L$ as the projection of $I$ onto $A_0 C_0$), we obtain the congruence of triangles $IKC$ and $ILA_0$ by the hypotenuse and an acute angle, from which it follows that $IK = IL$, so $L$ lies on $\omega$, and therefore $A_0 C_0$ is tangent to $\omega$.