Example 1 As shown in Figure 1, in $\triangle ABC$, $\angle BAC = \angle BCA = 44^{\circ}$, and $M$ is a point inside $\triangle ABC$ such that $\angle MCA = 30^{\circ}$ and $\angle MAC = 16^{\circ}$. Find the measure of $\angle BMC$.
(2005, Beijing Middle School Mathematics Competition (Grade 8))
As shown in Figure 1, draw $B D \perp A C$, with the foot of the perpendicular at $D$. Extend $C M$ to intersect $B D$ at point $O$, and connect $O A$.
It is easy to see that $B D$ is the axis of symmetry of $\triangle A B C$.
$$
\begin{array}{l}
\text { Then } \angle O A C=\angle M C A=30^{\circ} \\
\Rightarrow \angle M A C+\angle B A C=2 \angle O A C \\
\Rightarrow \angle B A O=\angle M A O . \\
\text { By } \angle A O D=90^{\circ}-\angle O A D \\
=60^{\circ}=\angle C O D,
\end{array}
$$

we get $\angle A O M=120^{\circ}=\angle A O B$.
Since $O A=O A$, then
$$
\triangle A B O \cong \triangle A M O \text {. }
$$

Thus, $O B=O M$.
Given $\angle B O M=120^{\circ}$, therefore,
$$
\angle O M B=\angle O B M=30^{\circ} \text {. }
$$

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