3. (7 points) In triangle $ABC$, angle bisectors are drawn from vertices $A$ and $B$, and a median is drawn from vertex $C$. It turns out that the points of their pairwise intersections form a right triangle. Find the angles of triangle $ABC$.


Solution. Let $M$ be the point of intersection of the angle bisectors of triangle $ABC$, and its median $CO$ intersects the drawn angle bisectors at points $K$ and $L$ (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_2b88941f3aaae556c0f7g-2.jpg?height=329&width=624&top_left_y=701&top_left_x=793)

Since $\angle AMB = 90^\circ + \frac{1}{2} \angle C > 90^\circ$, the angle at vertex $M$ in triangle $KLM$ cannot be a right angle. Therefore, from the problem's condition, it follows that $\angle AMB = 135^\circ$, so $\angle ACB = 90^\circ$. Consequently, $OC = OA = OB$, meaning triangles $AOC$ and $BOC$ are isosceles.

Without loss of generality, we can assume that the right angle in triangle $KLM$ is at vertex $K$. Then, in triangle $BOC$, the altitude $BK$ coincides with the angle bisector, so $OB = BC$. Thus, triangle $BOC$ is equilateral. Therefore, $\angle ABC = 60^\circ$, and hence $\angle BAC = 30^\circ$.

Answer. $90^\circ, 60^\circ, 30^\circ$.