Rozhnova $M$.

In triangle $ABC$, $O$ is the center of the circumscribed circle, and $H$ is the orthocenter. Through the midpoint of $OH$, a line parallel to $BC$ is drawn, intersecting sides $AB$ and $AC$ at points $D$ and $E$. It turns out that $O$ is the center of the inscribed circle of triangle $ADE$. Find the angles of triangle $ABC$.
From the condition, it follows that $A O$ is the bisector of angle $A$, which means $A B = A C$. Therefore, $O D H E$ is a rhombus.

First method. $\angle O D H = 2 \angle O D E = \angle B, \angle D O H = \angle D H O = 90^{\circ} - \angle B / 2 = \angle B H D$.

Let the line passing through $H$ parallel to $A C$ intersect $A B$ at point $K$. Since $\angle H K B = \angle A = \angle H O B$, points $H, O, K, B$ lie on the same circle. Since angle $K H B$ is a right angle, the center of this circle lies on line $A B$ and, therefore, coincides with point $D$ (see figure). Therefore, $\angle H B D = \angle B H D = 90^{\circ} - \angle B / 2$. On the other hand, this angle is equal to $\angle B - \angle A / 2$, from which the answer follows.

![](https://cdn.mathpix.com/cropped/2024_05_06_11215e94233515718ac4g-86.jpg?height=552&width=510&top_left_y=1528&top_left_x=774)

Second method. Let $P$ be the midpoint of segment $O H$. On the extension of $A B$ beyond point $B$, lay off segment $B F = B C$. Since $D O$ is the bisector of triangle $A D P$, then $A D: D P = A O: O P$. Therefore, $A B: B F = A B: B C = A D: 2 D P = A O: 2 O P = A O: O H$, which means $B O \parallel F H$. Since triangle $A O B$ is isosceles, triangle $A H F$ is also isosceles. Therefore, $C H$ is the perpendicular bisector of segment $A F$. Consequently, $A C = F C$, which means $\angle A = \angle B F C = \angle B C F = \angle B / 2$, from which the answer follows.

## Answer

$\angle A = 36^{\circ}, \angle B = \angle C = 72^{\circ}$.