[ Properties and characteristics of an isosceles triangle. ]

In triangle $A B C$, 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 isosceles triangle. Find the angles of triangle $A B C$.
Let $I$ be the point of intersection of the angle bisectors of triangle $ABC$, and the median $CO$ intersects the drawn bisectors at points $K$ and $L$ (see figure). Since

$\angle AIB = 90^\circ + \frac{1}{2} \angle C > 90^\circ$, the angle at vertex $I$ in the resulting triangle $KLI$ is $45^\circ$. Therefore, $\angle AIB = 135^\circ$, so $\angle ACB = 90^\circ$. Consequently, $OC = OA = OB$.

![](https://cdn.mathpix.com/cropped/2024_05_06_7adbd1f2e5e559c8684fg-03.jpg?height=412&width=849&top_left_y=1666&top_left_x=616)

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

## Answer

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