Shapovalov A.V.

In triangle $A B C$, a point $K$ is chosen on side $A B$, and the bisector $K E$ of triangle $A K C$ and the altitude $K H$ of triangle $B K C$ are drawn. It turns out that angle $E K H$ is a right angle. Find $B C$, if $H C=5$.
Since angles $A K C$ and $B K C$ are adjacent, their bisectors are perpendicular. Therefore, the bisector of angle $B K C$ coincides with $K H$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d1936174aeda27dcc10dg-08.jpg?height=620&width=618&top_left_y=-1&top_left_x=740)

Thus, in triangle $B K C$, segment $K H$ is both an altitude and a bisector. This means the triangle is isosceles, and $K H$ is its median to the base. Therefore, $B C = 2 H C = 10$.

## Answer

10.