3. On the hypotenuse $BC$ of the right triangle $ABC$, a point $K$ is chosen such that $AB = AK$. Segment $AK$ intersects the bisector $CL$ at its midpoint. Find the acute angles of triangle $ABC$. (I. Bogdanov)
Answer. $\angle B=54^{\circ}, \angle C=36^{\circ}$. Solution. Let the midpoint of the bisector $C L$ be $P$, and the angle $A B C$ be $\beta$; then $\angle A C L=\left(90^{\circ}-\beta\right) / 2$. In the right triangle $A C L$, the segment $A P$ is a median, so $A P=C P=L P$. Now from the isosceles triangles $A P L$ and $A B K$ we get $\angle A L P=\angle L A P=\angle B A K=180^{\circ}-2 \angle A B K=180^{\circ}-2 \beta$. On the other hand, $\angle A L P=\angle A B C+\angle L C B$ as an exterior angle in $\triangle B C L$. Therefore, $180^{\circ}-2 \beta=\beta+\left(90^{\circ}-\beta\right) / 2$, from which $5 \beta / 2=135^{\circ}$, that is, $\beta=54^{\circ}$. Then $\angle A C B=90^{\circ}-\angle A B C=36^{\circ}$.