Kazitsyna T.V.

In triangle $ABC$, points $X$ and $Y$ are taken on sides $AC$ and $BC$ such that $\angle ABX = \angle YAC$, $\angle AYB = \angle BXC$, and $XC = YB$. Find the angles of triangle $ABC$.
For the external angles $BXC$ and $AYB$ of triangles $ABX$ and $CAY$, we write the equalities $\angle BXC = \angle ABX + \angle BAX$, $\angle AYB = \angle CAY + \angle YCA$ (see the figure). Since by the condition $\angle BXC = \angle AYB$, $\angle ABX = \angle CAY$, then $\angle BAX = \angle YCA$, which means that triangle $ABC$ is isosceles, $AB = BC$.

For triangles $XBC$ and $YAB$, two sides and the angle not between them are equal: $\angle BXC = \angle AYB$, $XC = YB$, $BC = AB$. Such triangles are either equal or

$\angle XBC + \angle YAB = 180^\circ$ (we will prove this below), but the second case is impossible since $\angle XBC + \angle YAB < \angle ABC + \angle CAB = 180^\circ - \angle ACB < 180^\circ$. Therefore, triangles $XBC$ and $YAB$ are equal, and thus $\angle ABC = \angle BCA$, which means that triangle $ABC$ is equilateral.

Now we will prove the stated fact. Mark a point $B'$ on the ray $XB$ such that $XB' = YA$. Point $B'$ can be either inside or outside the segment $XB$ (see the figure).

Triangles $XB'C$ and $YAB$ are equal by two sides and the angle between them ($XC = YB$, $XB' = YA$, $\angle CXB' = \angle BYA$). Therefore, $CB' = CB$. If $B$ and $B'$ coincide, triangles $XBC$ and $YAB$ are equal. If $B'$ does not coincide with $B$, then in the isosceles triangle $B'CB$, the angles $B'BC$ and $BB'C$ are equal, and thus,

$\angle XBC + \angle XB'C = 180^\circ$, and $\angle XBC + \angle YAB = 180^\circ$.

## Answer

All angles are $60^\circ$.