Bogdanov I.I.

On the hypotenuse $AB$ of a right triangle $ABC$, a point $K$ is chosen such that $CK = BC$. The segment $CK$ intersects the bisector $AL$ at its midpoint.

Find the angles of triangle $ABC$.

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Let $\angle A=2 \alpha$, and $O$ be the intersection point of segments $C K$ and $A L$ (see the figure). Then $C O$ is the median to the hypotenuse of the right triangle $A C L$. Therefore,

$A O=O C=O L$, and $\angle O C A=\angle O A C=\angle O A K=\alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_06_60e93adedf73503f106cg-19.jpg?height=372&width=512&top_left_y=543&top_left_x=790)

Since triangle $C B K$ is isosceles, $\angle B=\angle B K C=\angle A C K+\angle K A C=3 \alpha$.

Thus, $2 \alpha+3 \alpha=\angle A+\angle B=90^{\circ}$. Therefore, $\alpha=18^{\circ}$. Consequently, $\angle B=3 \cdot 18^{\circ}=54^{\circ}$, and $\angle A=2 \cdot 18^{\circ}=36^{\circ}$.

## Answer

$36^{\circ}, 54^{\circ}, 90^{\circ}$.