5. Inside an isosceles triangle $A B C (A B=A C)$, a point $K$ is marked. Point $L$ is the midpoint of segment $B K$. It turns out that $\angle A K B=\angle A L C=90^{\circ}, A K=C L$. Find the angles of triangle $A B C$.
Answer: the triangle is equilateral. Note that $\triangle A L C=\triangle B K A$ by the leg and hypotenuse. Therefore, $A L=B K=2 K L$, so $\angle K A L=30$. In addition, the sum of angles $B A K$ and $C A L$ in these triangles is 90, so $\angle B A C=90-30-60$.