Nistoroscu $K$.

Find the angles of an acute-angled triangle $ABC$ if it is known that its bisector $AD$ is equal to the side $AC$ and is perpendicular to the segment $OH$, where $O$ is the center of the circumscribed circle, and $H$ is the orthocenter of triangle $ABC$.
Continue the segment $C H$ until it intersects the circumscribed circle of the given triangle at point $F$ and prove that triangle $A O F$ is equilateral.

## Solution

Extend the segments $A O$ and $A H$ until they intersect the circumscribed circle of triangle $A B C$ at points $E$ and $M$ respectively. Since $\angle A M E=90^{\circ}$, it follows that $E M \| B C$. Therefore, the smaller arcs $M C$ and $E B$ are equal. Consequently, $\angle C A M = \angle B A E$, and since $A D$ is the angle bisector of $\angle B A C$, we have $\angle O A D = \angle H A D$. Thus, the altitude $A K$ of triangle $O A H$ is also its angle bisector. Therefore, triangle $O A H$ is isosceles.

Moreover, the altitude $A P$ of the isosceles triangle $C A D$ is also its angle bisector. Therefore,

$$
\angle C A M = \angle M A D = \angle D A O = \angle O A B
$$

Let each of these angles be denoted by $\alpha$. Extend the segment $C H$ until it intersects the circle at point $F$. Then

$$
\angle B A F = \angle B C F = \angle B A H
$$

Thus, the altitude $A L$ of triangle $H A F$ is also its angle bisector. Therefore, triangle $H A F$ is isosceles, and $A L$ is its median. Consequently, if $R$ is the radius of the circle, then

$$
A F = A H = A O = R = F O
$$

i.e., triangle $A O F$ is equilateral. Therefore,

$$
\angle A O F = 60^{\circ}, \angle A C F = 30^{\circ},
$$

and since $\angle A C F = 90^{\circ} - 4 \alpha$, we have $90^{\circ} - 4 \alpha = 30^{\circ}$. Solving for $\alpha$, we get $\alpha = 15^{\circ}$. Therefore,

$$
\angle B A C = 4 \alpha = 60^{\circ}, \angle A C B = 90^{\circ} - \angle C A P = 90^{\circ} - \alpha = 75^{\circ}, \angle A B C = 45^{\circ}.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_6741038a32a996d2ce33g-08.jpg?height=452&width=1089&top_left_y=0&top_left_x=490)

## Answer

$60^{\circ}, 45^{\circ}, 75^{\circ}$.