3. Circle $\omega_{1}$ with center $O$ intersects circle $\omega_{2}$, which passes through point $O$, at points $K$ and $L$. A line through point $O$ intersects circle $\omega_{2}$ again at point $A$. Segment $OA$ intersects circle $\omega_{1}$ at point $B$. Find the ratio of the distances from point $B$ to the lines $A L$ and $K L$.
Answer: $1: 1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c43b1f7fb831d10d42ecg-17.jpg?height=417&width=651&top_left_y=268&top_left_x=722)

Solution. Notice that $\angle O A K = \angle O A L$ as angles subtended by equal chords $O K$ and $O L$. Therefore, $A B$ is the angle bisector of $\angle L A K$. Moreover, $\angle A K L = \angle A O L$ as angles subtended by the arc $A L$. Then,

$$
\angle B K L = \frac{1}{2} \angle B O L = \frac{1}{2} \angle A K L
$$

Thus, $K B$ is the angle bisector of $\angle A K L$. Therefore, point $B$ is the incenter of triangle $A K L$. This means that it is equidistant from the lines $A L$ and $K L$.