8. Variant 1. Point $I$ is the center of the circle inscribed in triangle $A B C$ with sides $B C=6$, $C A=8, A B=10$. Line $B I$ intersects side $A C$ at point $K$. Let $K H$ be the perpendicular dropped from point $K$ to side $A B$. Find the distance from point $I$ to line $K H$.
Answer: 2.

Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_63606fabfe4f054b870eg-7.jpg?height=504&width=631&top_left_y=220&top_left_x=736)

By the converse of the Pythagorean theorem, angle $C$ is a right angle. Then, triangles $B K C$ and $B K H$ are congruent by the hypotenuse and an acute angle. Therefore, these triangles are symmetric with respect to the line $B K$. Thus, the distance from $I$ to $K H$ is equal to the distance from $I$ to $C K$, i.e., the radius of the inscribed circle in triangle $A B C$. For a right triangle, this radius is given by $\frac{A C + C B - B A}{2} = 2$.