Four, (15 points) As shown in Figure 4, in $\triangle A B C$, $\angle B A C$ $=\angle B C A=44^{\circ}, M$ 
is a point inside $\triangle A B C$, 
such that $\angle M C A=30^{\circ}$, $\angle M A C=16^{\circ}$. Find the measure of $\angle B M C$.
In $\triangle A B C$, from $\angle B A C=\angle B C A=44^{\circ}$, we get
$$
A B=B C, \angle A B C=92^{\circ} \text {. }
$$

As shown in Figure 6, draw $B D \perp$
$A C$ at point $D$, extend $C M$
to intersect $B D$ at point $O$, and connect
$O A$. Then we have
$$
\begin{array}{l}
\angle O A C=\angle M C A \\
=30^{\circ}, \\
\angle B A O=\angle B A C-\angle O A C=44^{\circ}-30^{\circ}=14^{\circ}, \\
\angle O A M=\angle O A C-\angle M A C=30^{\circ}-16^{\circ}=14^{\circ} .
\end{array}
$$

Therefore, $\angle B A O=\angle M A O$.
$$
\text { Also, } \angle A O D=90^{\circ}-\angle O A D=90^{\circ}-30^{\circ}=60^{\circ}=\angle C O D \text {, }
$$

Therefore, $\angle A O M=120^{\circ}=\angle A O B$.
Since $A O=A O$, thus, $\triangle A B O \cong \triangle A M O$.
Hence $O B=O M$.
Given $\angle B O M=120^{\circ}$, therefore,
$$
\angle O M B=\angle O B M=\frac{180^{\circ}-\angle B O M}{2}=30^{\circ} \text {. }
$$

Therefore, $\angle B M C=180^{\circ}-\angle O M B=150^{\circ}$.