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

![](https://cdn.mathpix.com/cropped/2024_05_06_c43b1f7fb831d10d42ecg-19.jpg?height=468&width=736&top_left_y=1999&top_left_x=680)

Solution. Triangles $O_{1} A K$ and $O_{2} A L$ are isosceles and have a common base angle. 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 $\angle A D L = \angle A C K = \alpha$. Since $K L \| C D$, we also get $\angle K L N = \angle L K N = \alpha$. Then

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

Therefore, the quadrilateral $L N K O_{2}$ is cyclic, from which we get

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

This means that the line $O_{2} N$ is the angle bisector of the vertex angle in the isosceles triangle $A O_{2} L$, and therefore also the altitude. Thus, the lines $O_{2} N$ and $O_{1} A$ are perpendicular.