B2.

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Let $S$ be the center of the circle $\mathcal{K}$ and $T$ the intersection of the line $M S$ with the side $A B$. Since the trapezoid $A B C D$ is cyclic, it is isosceles with legs $A D$ and $B C$. Let $\alpha = \angle B A D = \angle C B A$. Due to symmetry, we can assume that $|A B| > |C D|$. The triangle $B M A$ is isosceles with the vertex at $M$, and the line $M S$ is the altitude of this triangle due to symmetry, hence it is perpendicular to the side $A B$. The angle $\angle B S D$ is the central angle over the arc $\widehat{B D}$ of the circle $\mathcal{K}$, and the angle $\angle B A D = \alpha$ is the inscribed angle over the same arc, so $\angle B S D = 2 \alpha$. Since triangles $S N D$ and $S N B$ are congruent, it follows that $\angle N S D = \angle B S N = \alpha$. Therefore, triangles $A T M$ and $S D N$ are similar, as they share two angles (a right angle and the angle $\alpha$), so $\angle A M T = \angle S N D$. By the inscribed angle theorem, it follows that points $D, S, N$, and $M$ are concyclic, hence $\angle S M N = \angle S D N = \frac{\pi}{2}$. This proves that the line $M S$ is perpendicular to the segment $M N$ and $A B$, so these two segments are parallel.

2. method. Let $S$ be the center of the circle $\mathcal{K}$ and $T$ the intersection of the line $M S$ with the side $A B$. Similarly to the first solution, we conclude that the trapezoid $A B C D$ is isosceles and denote $\alpha = \angle B A D = \angle C B A$. Due to symmetry, we can again assume that $|A B| > |C D|$ or $\alpha < \frac{\pi}{2}$. Then $\angle D M B = \angle A M B = \pi - \angle M B A - \angle B A M = \pi - 2 \alpha$. The angle $\angle B S D$ is the central angle over the arc $\widehat{B D}$ of the circle $\mathcal{K}$, and the angle $\angle B A D = \alpha$ is the inscribed angle over the same arc, so $\angle B S D = 2 \alpha$. The quadrilateral $S B N D$ is cyclic by Thales' theorem, so $\angle D N B = \pi - \angle B S D = \pi - 2 \alpha$. This shows that $\angle D N B = \angle D M B$, so the quadrilateral $D B N M$ is also cyclic. From this it follows that

$$\angle A M N = \angle D M N = \pi - \angle N B D = \pi - \left(\frac{\pi}{2} - \angle D B S\right) = \frac{\pi}{2} + \angle D B S = \frac{\pi}{2} + \angle D M S$$

Due to symmetry, the line $M S$ or $M T$ is the altitude of the isosceles triangle $B M A$, so $\angle M T A = \frac{\pi}{2}$ and $\angle D M S = \angle D M T = \frac{\pi}{2} - \angle T A M = \frac{\pi}{2} - \alpha$. From the above equality, it follows that

$$\angle A M N = \frac{\pi}{2} + \angle D M S = \frac{\pi}{2} + \left(\frac{\pi}{2} - \alpha\right) = \pi - \alpha = \angle A D C$$

Therefore, the segments $M N$ and $A B$ are parallel.

1. ... method:
Observation that the trapezoid $A B C D$ or the triangle $A B M$ is isosceles ... 1 point.
Observation that $M S$ is the altitude of the triangle $A B M$ ... 1 point.
Proof that $\angle B S D = 2 \angle B A D$ ... 1 point.
Proof that the angles $\angle D S B$ and $\angle N S B$ are equal or that the triangles $A T M$ and $S D M$ are similar ... 2 points.
Application of cyclic properties and conclusion ... 2 points.
Proof that $\angle D M B = \pi / 2 - 2 \alpha$ ... 1 point.
Proof that $\angle B S D = 2 \angle B A D$ ... 1 point.
Proof that $\angle B N D = \pi - \angle B S D = \pi - 2 \alpha$ or $\angle B N D = \angle D M B$ ... 1 point.
Observation that $M S$ or $M T$ is the altitude of the triangle $A M B$ ... 1 point.
Proof that $\angle D M S = \angle D M T = \pi / 2 - \alpha$ ... 1 point.
Proof that $\angle A M N = \angle A D C$ ... 1 point.