3. A circle $\omega$ is circumscribed around triangle $A B C$. Tangents to the circle, drawn at points $A$ and $B$, intersect at point $K$. Point $M$ is the midpoint of side $A C$. A line passing through point $K$ parallel to $A C$ intersects side $B C$ at point $L$. Find the angle $A M L$.
Answer: $90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-04.jpg?height=477&width=489&top_left_y=1857&top_left_x=795)

Solution. Let $\alpha=\angle A C B$. The angle between the tangent $A K$ and the chord $A B$ of the circle $\omega$ is equal to the inscribed angle that subtends $A B$, hence $\alpha=\angle B A K=\angle A B K$. Since
$A C \| K L$, we get $\angle B L K=\angle A C B=\angle B A K$. Therefore, the quadrilateral $A L B K$ is cyclic. Then

$$
\angle A L C=180^{\circ}-\angle A L B=\angle A K B=180^{\circ}-2 \alpha \quad \text { and } \quad \angle L C A=\alpha=180^{\circ}-\angle A L C-\alpha=\angle L A C .
$$

Thus, triangle $A L C$ is isosceles, and its median $L M$ is also an altitude. Therefore, $\angle A M L=90^{\circ}$.