In triangle $A B C$, points $X$ and $Y$ are taken on sides $A C$ and $B C$ respectively, such that $\angle A B X=\angle Y A C, \angle A Y B=$ $\angle B X C, X C=Y B$.

Find the angles of triangle $A B C$.
Lemma. If the sides $K_{1} L_{1}, K_{1} M_{1}$ and the angle $K_{1} L_{1} M_{1}$ of triangle $K_{1} L_{1} M_{1}$ are respectively equal to the sides $K L$, $K M$ and the angle $K L M$ of triangle $K L M$, then either $\angle K_{1} M_{1} L_{1}=\angle K M L$ and in this case triangle $K_{1} L_{1} M_{1}$ is equal to triangle $K L M$, or $\angle K_{1} M_{1} L_{1}=180^{\circ}-\angle K M L$.

Proof. Let triangle $K L M$ be given (see figure). Construct triangle $K_{1} L_{1} M_{1}$ using the given sides $K_{1} L_{1}=K L, K_{1} M_{1}=K_{1} L_{1}$ and the angle $K_{1} L_{1} M_{1}$, equal to the angle $K L M$. For this, from the beginning of an arbitrary ray, we lay off the segment $L_{1} K_{1}=L K$, then in one of the half-planes into which the line $L_{1} K_{1}$ divides the plane, we lay off the ray $L_{1} P$, forming with the ray $L_{1} K_{1}$ an angle equal to the given angle $KML$. With the center at point $K_{1}$, we construct a circle with radius equal to $KM$. If this circle does not intersect the constructed ray, then the problem has no solutions; if the circle has a unique common point with the ray, then the problem has one solution; if the circle intersects the ray at two points $M_{1}$ and $M_{2}$ (point $M_{2}$ lies between $L_{1}$ and $M_{1}$), then triangle $M_{1} K_{1} M_{2}$ is isosceles, therefore $\angle K_{1} M_{2} L_{1}=\angle K_{1} M_{2} M_{1}=\angle K_{1} M_{1} M_{2}=180^{\circ}-\angle K_{1} M_{1} L_{1}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-15.jpg?height=894&width=826&top_left_y=319&top_left_x=613)

Let's return to our problem. Denote $\angle A B X=\angle Y A C=\alpha, \angle A Y B=\angle B X C=\beta$ (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-15.jpg?height=566&width=608&top_left_y=1281&top_left_x=725)

Since $A Y B$ and $B X C$ are exterior angles of triangles $A Y C$ and $A X B$, we have $\angle C=\angle A C Y=\angle A Y B-\angle Y A C=\beta-\alpha=$ $\angle B X C-A B X=\angle B A X=\angle A$.

Thus, triangle $A B C$ is isosceles, $A B=B C$. Consider triangles $X B C$ and $Y A B$. It is known that $B C=A B, C X=Y B$ and $\angle B X C=\angle A Y B$. From the lemma, it follows that either triangles $X B C$ and $Y A B$ are equal, or $\angle X B C + \angle Y A B=180^{\circ}$. The second option is impossible, as in this case the sum of the angles of triangle $A Y B$ would be greater than $180^{\circ}$. From the equality of triangles $X B C$ and $Y A B$, it follows that the angles $C$ and $B$ are equal, hence triangle $A B C$ is equilateral.

## Answer

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