10.3. Given a right triangle $A B C$. A point $M$ is taken on the hypotenuse $A C$. Let $K, L$ be the centers of the circles inscribed in triangles $A B M$ and $C B M$ respectively. Find the distance from point $M$ to the midpoint of $K L$, if the radius $R$ of the circle circumscribed around triangle $B K L$ is known.
Answer: $\frac{R \sqrt{2}}{2}$. Solution. Note that angle $K M L$ is a right angle, since $\angle K M L=\angle K M B+\angle L M B=$ $=\frac{1}{2}(\angle A M B+\angle C M B)=90^{\circ}$. Similarly, we obtain that $\angle K B L=\frac{1}{2} \cdot 90^{\circ}=45^{\circ}$. Then in triangle $B K L$ we have $K L=2 R \cdot \sin 45^{\circ}=R \sqrt{2}$. The hypotenuse $K L$ in the right triangle $K M L$ is twice the desired median, from which the result $\frac{R \sqrt{2}}{2}$ follows.