1.2. In triangle $ABC$ with angle $BAC$ equal to $24^{\circ}$, points $X$ and $Y$ are taken on sides $AB$ and $AC$ respectively. The circle centered at $Y$ and passing through $A$ also passes through $X$, and the circle centered at $X$ and passing through $B$ also passes through $C$ and $Y$. Find $\angle ABC$.
1.2. Answer: $54^{\circ}$.

Note that $A Y=X Y$, since $X$ lies on the circle centered at $Y$ passing through $A$. (see Fig. 9).

Then $\triangle A Y X$ is isosceles, and $\angle A X Y=\angle X A Y=24^{\circ}$. Since $\angle X Y C$ is the exterior angle for triangle $A X Y$, $\angle X Y C=\angle A X Y+\angle X A Y=48^{\circ}$. Points $Y, C, B$ lie on the circle centered at $X$, so $X Y=X C=X B$, and triangles $Y X C$ and $B X C$ are isosceles. Therefore, $\angle X C Y=\angle X Y C=48^{\circ}, \angle X C B=\angle A B C$. The sum of the angles in triangle $A B C$ should be $180^{\circ}$, so $180^{\circ}=\angle B A C+\angle X C Y+\angle X C B+\angle A B C=24^{\circ}+48^{\circ}+2 \angle A B C$. From this, we get that $2 \angle A B C=180^{\circ}-24^{\circ}-48^{\circ}=108^{\circ}$, thus $\angle A B C=54^{\circ}$.