3. Circles $\omega_{1}$ and $\omega_{2}$ with centers $O_{1}$ and $O_{2}$ respectively intersect at point $B$. The extension of segment $O_{2} B$ beyond point $B$ intersects circle $\omega_{1}$ at point $K$, and the extension of segment $O_{1} B$ beyond point $B$ intersects circle $\omega_{2}$ at point $L$. The line passing through point $B$ parallel to $K L$ intersects circles $\omega_{1}$ and $\omega_{2}$ again at points $A$ and $C$ respectively. The rays $A K$ and $C L$ intersect at point $N$. Find the angle between the lines $O_{1} N$ and $O_{2} B$.
Answer: $90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c43b1f7fb831d10d42ecg-23.jpg?height=483&width=662&top_left_y=267&top_left_x=731)

Solution. Since $\angle O_{1} B K = \angle O_{2} B L$, the isosceles triangles $O_{1} B K$ and $O_{2} B L$ are similar. Therefore, their vertex angles are equal. Let the common value of these angles be $2 \alpha$. Since an inscribed angle is half the corresponding central angle, we have $\angle B C L = \angle B A K = \alpha$. Since $K L \| A C$, we also get $\angle K L N = \angle L K N = \alpha$. Then,

$$
\angle L N K = 180^{\circ} - 2 \angle L K N = 180^{\circ} - 2 \alpha = 180^{\circ} - \angle B O_{1} K = 180^{\circ} - \angle L O_{1} K
$$

Therefore, the quadrilateral $L N K O_{1}$ is cyclic, from which we get

$$
\angle L O_{1} N = \angle L K N = \alpha = \frac{1}{2} \angle B O_{1} K
$$

Thus, the line $O_{1} N$ is the angle bisector of the vertex angle in the isosceles triangle $B O_{1} K$, and therefore also the altitude. Hence, the lines $O_{1} N$ and $O_{2} B$ are perpendicular.