Answer: $40^{\circ}$. Solution: Extend the triangle to form a parallelogram $A C B D$. Then $M$ is the midpoint of segment $C D$.

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Since $A P = B C = A D$ and $B Q = A C = B D$, triangles $A P D$ and $B Q D$ are isosceles. Therefore, $\angle Q D P = \angle A D P + \angle B D Q - \angle A D B = \left(90^{\circ} - \frac{\angle D A B}{2}\right) + \left(90^{\circ} - \frac{\angle D B A}{2}\right) - 100^{\circ} = 80^{\circ} - \frac{(\angle D A B + \angle D B A)}{2} = 40^{\circ}$. It remains to note that $\angle Q D P = \angle K M N$, since $M K$ and $M N$ are the midlines of triangles $D Q C$ and $D P C$ respectively.
2. On the side AB of triangle ABC with an angle of $100^{\circ}$ at vertex C, points $P$ and $Q$ are taken such that $A P=B C$ and $B Q=A C$. Let $M, N, K-$ be the midpoints of segments $A B, C P, C Q$ respectively. Find the angle NМК. (M. Kungozhin + jury)