2. In $\triangle A B C$, it is known that $\angle C=100^{\circ}$, points $P$ and $Q$ are both on side $A B$, such that $A P=B C, B Q=A C$, the midpoints of segments $A B$, $C P$, and $C Q$ are $M$, $N$, and $K$ respectively. Find $\angle N M K$.
2. As shown in Figure 1, extend $\triangle A B C$ to form $\triangle C B D$.

Thus, $M$ is the midpoint of line segment $C D$.
Since $A P = B C = A D, B Q = A C = B D$, therefore,
$$
\begin{array}{c}
\angle Q D P = \angle A D P + \angle B D Q - \angle A D B \\
= \frac{180^{\circ} - \angle D A B}{2} + \frac{180^{\circ} - \angle D B A}{2} - 100^{\circ} \\
= 80^{\circ} - \frac{\angle D A B + \angle D B A}{2} = 40^{\circ}.
\end{array}
$$

Next, we prove: $\angle N M K = \angle Q D P$.
Since line segments $M K$ and $M N$ are the midlines of $\triangle C Q D$ and $\triangle C P D$, respectively, we have,
$$
M K \parallel D Q, M N \parallel D P.
$$

Thus, $\angle N M K = \angle Q D P = 40^{\circ}$.