27. Points $X$ and $Y$ are the midpoints of arcs $A B$ and $B C$ of the circumcircle of triangle $A B C$. $B L$ is the angle bisector of this triangle. It turns out that $\angle A B C=2 \angle A C B$ and $\angle X L Y=90^{\circ}$. Find the angles of triangle $A B C$.
27. Answer: $\angle B=90^{\circ}, \angle A=\angle C=45^{\circ}$.

By the condition $\angle C=\angle B / 2=\angle L B C$, so triangle $B L C$ is isosceles: $B L=C L$. In addition, $B Y=Y C$, so triangles $B L Y$ and $C L Y$ are equal by three sides, and $L Y-$ is the bisector of angle $B L C$. By the condition, line $L X$ is perpendicular to this bisector, so $L X-$ is the bisector of angle $B L A$.

Now consider triangles $X A L$ and $X B L$. They have a common side $L X$, equal sides $A X=B X$, and equal angles $A L X$ and $B L X$. Unfortunately, these angles are not between the equal sides, so the first criterion for equality is not applicable. However, this means that angles $X A L$ and $X B L$ are either equal or supplementary to $180^{\circ}$. (It is easiest to explain this using the sine theorem in triangles $X A L$ and $X B L: \sin X A L=\sin X B L$.) In the second case, quadrilateral $A X B L$ would be cyclic, which is impossible: point $L$ does not lie on the circle passing through $A, B$ and $X$. Therefore, triangles $X A L$ and $X B L$ are indeed equal and $A L=C L$.

Thus, $A L=B L=C L$, i.e., the median of triangle $A B C$ is equal to half the side $A C$. This means that the triangle is right-angled. And since the median $B L$ coincides with the bisector, it is also isosceles.