Bogdanov I.I.

In a right triangle $ABC$, a point $K$ is chosen on the hypotenuse $BC$ such that $AB = AK$. The segment $AK$ intersects the bisector $CL$ at its midpoint.

Find the acute angles of triangle $ABC$.
Let the midpoint of the bisector $C L$ be $P$, and the angle $A B C$ be $\beta$; then $\angle A C L=1 / 2\left(90^{\circ}-\beta\right)$. In the right triangle $A C L$, the segment $A P$ is a median, so $A P=C P=L P$. 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, $180^{\circ}-2 \beta=\beta+1 / 2$ $\left(90^{\circ}-\beta\right)$, from which $5 \beta / 2=135^{\circ}$, that is,

$\beta=54^{\circ}$. Then $\angle A C B=90^{\circ}-\angle A B C=36^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ccb5665c9ecd710ea159g-25.jpg?height=400&width=529&top_left_y=180&top_left_x=770)

## Answer

$\angle B=54^{\circ}, \angle C=36^{\circ}$.