8,9 $[\quad$ Inscribed angle, based on the diameter

In triangle $A B C$, it is known that $\angle B A C=\alpha, A C=b$. The inscribed circle touches the sides $A B$ and $B C$ at points $M$ and $N$, the bisector of angle $B A C$ intersects the line $M N$ at point $K$. Find the distance from point $K$ to the line $A C$.
Prove that $\angle A K C=90^{\circ}$.

## Solution

Let $\angle A C B=\gamma, \angle A B C=\beta$. From the isosceles triangle $M B N$, we find that

$$
\angle B M N=\angle B N M=90^{\circ}-\frac{\beta}{2}=90^{\circ}-\left(90^{\circ}-\frac{\alpha}{2}-\frac{\gamma}{2}\right)=\frac{\alpha}{2}+\frac{\gamma}{2}.
$$

$$
\angle A K M=\angle B M K-\angle M A K=\frac{\alpha}{2}+\frac{\gamma}{2}-\frac{\alpha}{2}=\frac{\gamma}{2}
$$

Let $O$ be the center of the inscribed circle of triangle $A B C$. From the above, it follows that the segment $O N$ is seen from points $K$ and $C$ at the same angle $\frac{\gamma}{2}$. Therefore, points $K, C, N$, and $O$ lie on the same circle, and since $\angle O N C=90^{\circ}$, $O C$ is the diameter of this circle. Consequently, $\angle A K C=\angle O K C=90^{\circ}$.

Thus, $A C$ is the hypotenuse of the right triangle $A K C$, and the required distance is the height $K H$ of this triangle, dropped to the hypotenuse. Therefore,

$$
K H=A K \cdot \sin \angle C A K=A C \cdot \cos \angle C A K \cdot \sin \angle C A K=b \cos \frac{\alpha}{2} \cdot \sin \frac{\alpha}{2}=\frac{1}{2} b \sin \alpha
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_f3f7c9453eb6f3f37ea7g-18.jpg?height=583&width=806&top_left_y=592&top_left_x=635)

## Answer

$\frac{b}{2} \sin \alpha$.