4. In a right triangle $ABC$ with angle $\angle A=60^{\circ}$, a point $N$ is marked on the hypotenuse $AB$, and the midpoint of segment $CN$ is marked as point $K$. It turns out that $AK=AC$. The medians of triangle $BCN$ intersect at point $M$. Find the angle between the lines $AM$ and $CN$.
Solution. By the property of a right triangle with an angle of $30^{\circ}$, we get that $A B: A C=2: 1$. Therefore, $A B: A K=2: 1$. By the property of medians in a triangle, $\mathrm{BM}: \mathrm{MK}=2: 1$. Then, by the converse of the angle bisector theorem, we get that $A M$ is the bisector of angle $\angle \mathrm{BAK}$. Let $\angle \mathrm{B}=\bar{\alpha}=$. Then $\angle \mathrm{K} M=\alpha$,

![](https://cdn.mathpix.com/cropped/2024_05_06_a7b664180bdae74858d0g-2.jpg?height=371&width=662&top_left_y=720&top_left_x=1254)
$\angle \mathrm{KAC}=60^{\circ}-2 \alpha$. Since triangle KAC is isosceles, $\angle \mathrm{KCA}=60^{\circ}+\alpha$. By the theorem of the sum of angles in a triangle, we get that the angle between lines $A M$ and $C N$ is $180^{\circ}-\left(60^{\circ}-\alpha+60^{\circ}+\alpha\right)=60^{\circ}$.

Criteria. If the solution is incorrect - 0 points. If it is proven that triangle OXY is isosceles - 3 points. If the solution is correct - 7 points.