In a circle with center $O$, a trapezoid $K L M N$ is inscribed, where $K L$ is parallel to $M N$, $K L=8$, $M N=2$, and the angle $N K L$ is $45^{\circ}$. Chord $M A$ of the circle intersects segment $K L$ at point $B$, such that $K B=3$. Find the distance from point $O$ to the line $A K$.
$N B$ - the height of the trapezoid.

## Solution

Let $P$ be the projection of vertex $N$ onto the base $K L$ of the trapezoid $K L M N$. Since the trapezoid is inscribed in a circle, it is isosceles. Therefore,

$$
K P=\frac{K L-M N}{2}=\frac{6}{2}=3
$$

Thus, point $P$ coincides with point $B$, and $N B$ is the height of the trapezoid. Then

$$
\begin{gathered}
M L=K N=\frac{K B}{\cos 45^{\circ}}=3 \sqrt{2} \\
B M=\sqrt{M N^{2}+N B^{2}}=\sqrt{4+9}=\sqrt{13}
\end{gathered}
$$

From the similarity of triangles $K B A$ and $M B L$, it follows that $\frac{K A}{M L}=\frac{K B}{B M}$. From this, we find that

$$
K A=M L \cdot \frac{K B}{B M}=\frac{9 \sqrt{2}}{\sqrt{13}}
$$

Let $R$ be the radius of the circle. Then

$$
O A=O K=R=\frac{L N}{2 \sin 45^{\circ}}=\frac{\sqrt{N B^{2}+B L^{2}}}{\sqrt{2}}=\sqrt{17}
$$

Let $B H$ be the height of the isosceles triangle $A O K$. Then

$$
O H=\sqrt{O A^{2}-A H^{2}}=\sqrt{R^{2}-\frac{A K^{2}}{4}}=\frac{19}{\sqrt{26}}
$$

## Answer

$\frac{19}{\sqrt{26}}$.