8. In $\triangle A B C$, $\angle A B C=100^{\circ}, \angle A C B$ $=65^{\circ}$, points $M, N$ are taken on sides $A B, A C$ respectively, such that $\angle M C B=55^{\circ}, \angle N B C=80^{\circ}$. Determine the measure of $\angle N M C$.
8. $25^{\circ}$.

It is easy to know that $\angle B A C=15^{\circ}$.
As shown in Figure 7, construct the circumcircle of $\triangle M C B$ intersecting the extension of $B N$ at point $M_{1}$. In this circle, the arcs $\overparen{C M_{1}}$ and $\overparen{C M}$ subtend complementary angles. Therefore, $C M_{1}=C M$.
$$
\begin{array}{l}
\text { Also, } \angle M_{1} C M \\
=\angle M_{1} B M \\
=100^{\circ}-80^{\circ}=20^{\circ}, \\
\angle A C M \\
=65^{\circ}-55^{\circ}=10^{\circ},
\end{array}
$$

Thus, $\angle M_{1} C N=10^{\circ}$.
Since $C N=C N$, then $\triangle M_{1} C N \cong \triangle M C N$.
Therefore, $\angle N M C=\angle N M_{1} C=\angle C M B$
$$
=\angle B A C+\angle A C M=15^{\circ}+10^{\circ}=25^{\circ} \text {. }
$$
(Li Yanlin provided)