Answer: $\sqrt{r(a+r)}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9304f7a3bc6f36cb1f9ag-25.jpg?height=508&width=465&top_left_y=494&top_left_x=881)

Solution. Let $L$ be the intersection point of $M N$ with the ray $B O$. Note that triangle $M B N$ is isosceles, and $B L$ is the bisector of its angle $B$. Therefore, $B L$ is also the altitude and median of triangle $M B N$. Then $\angle K L M=90^{\circ}$, and the right triangles $K L M$ and $B L N$ are congruent, from which $M K=B N$. Moreover,

$$
\angle M K L=\angle L B C=\frac{1}{2} \angle A B C=\angle M T K
$$

Thus, triangles $M K T$ and $M L K$ are similar by two angles. Therefore, $\angle M K T=90^{\circ}$, and triangle $M K T$ is also similar to $O N B$. Then

$$
\frac{K T}{M K}=\frac{B N}{O N} \Longleftrightarrow \frac{a}{B N}=\frac{B N}{r} \Longleftrightarrow B N^{2}=a r
$$

By the Pythagorean theorem,

$$
B O=\sqrt{B N^{2}+O N^{2}}=\sqrt{a r+r^{2}}=\sqrt{r(a+r)} .
$$
3. A circle with center $O$ and radius $r$ is tangent to the sides $BA$ and $BC$ of the acute angle $ABC$ at points $M$ and $N$ respectively. A line passing through point $M$ parallel to $BC$ intersects the ray $BO$ at point $K$. On the ray $MN$, a point $T$ is chosen such that $\angle MTK = \frac{1}{2} \angle ABC$. Find the length of the segment $BO$, if $KT = a$.