Answer: $60^{\circ}$

Solution. Note that point $O$ lies on the perpendicular bisector of segment $B C$. On the other hand, point $O$ lies on the bisector of angle $A$. However, these two lines intersect at a point lying on the circumcircle of triangle $A B C$, and dividing the arc $B C$ in half. Therefore, point $O$ lies on the circumcircle of triangle $A B C$.

The angles at the vertices of triangle $A B C$ will be denoted by the same letter. Consider the isosceles triangle $O H C, O H=O C$. Then, from the observation about the location of point $O$ and the properties of inscribed angles, we immediately get that the angles at the base of this isosceles triangle are equal to $90^{\circ}-B+A / 2$. Similarly, the angle at the base of the isosceles triangle $O H B(O H=$ $O B)$ is equal to $90^{\circ}-C+A / 2$. But the sum of these two angles is equal to the angle $B H C$, which, in turn, is equal to the angle $180^{\circ}-A$ (by the theorem of vertical angles). Therefore, $180^{\circ}-B-C+A=180^{\circ}-A$, or $3 A=180$.

## Criteria.

(-.) The answer is incorrect and there is no complete solution, but there are significant advances in the solution.

(-/+) The correct answer is obtained, but the solution uses without proof that point $O$ belongs to the circumcircle of $\triangle A B C$.

(+/-) It is proven that point $O$ lies on the circumcircle of $\triangle A B C$.

(+.) The answer is incorrect due to an arithmetic error.

$(+)$ Correct answer + full justification.
8.6. The altitudes of an acute-angled scalene triangle $ABC$ intersect at point $H$. $I$ is the incenter of triangle $ABC$, $O$ is the circumcenter of triangle $BHC$. It is known that point $I$ lies on the segment $OA$. Find the angle $BAC$.