Three, (30 points) In $\triangle A B C$, $A B=A C$, point $M$ is on $A B$ and $M A=M C$, point $N$ is on $A C$ and $C N=C B, \angle A: \angle N B A=2: 3$. Find the degree measure of $\angle N M C$.
Let $\angle A=\alpha$, then $\angle N B A=\frac{3}{2} \alpha, \angle B N C=\frac{5}{2} \alpha$, $\angle C B i V=\frac{5}{2} \alpha, \angle C B A=\frac{5}{2} \alpha+\frac{3}{2} \alpha=4 \alpha$.
$\angle M C A=20^{\circ}, \angle M C B=60^{\circ}$.
Draw $M K / / B C$. Then quadrilateral $B M K C$ is an isosceles trapezoid. Extend $K B$ to intersect $M C$ at $O$, and connect $O N$. Since $\angle M C B=\angle O B C=60^{\circ}$, $\triangle O B C, \triangle O M K$ are equilateral triangles, $B K=C M=$ $M A$, we have
$$
\begin{array}{l}
\angle K B A=\angle A=20^{\circ}, \\
\angle C K B=20^{\circ}+20^{\circ}=40^{\circ} .
\end{array}
$$

Also, since $C B=C O=C N, \angle A C B=20^{\circ}$, we have $\angle C O N=$ $80^{\circ}$. Also, since $\angle K O C=120^{\circ}$, we have $\angle K O N=40^{\circ}$. Thus, $N K=$ $N O$. Therefore, $\triangle M K N \cong \triangle M O N$. So, $\angle K M O=30^{\circ}$.