Example 4 As shown in Figure 6, in $\triangle A B C$, it is known that $\angle B A C$ $=\angle B C A=44^{\circ}, M$ is a point inside $\triangle A B C$, such that $\angle M C A=30^{\circ}$, $\angle M A C=16^{\circ}$. Find the degree measure of $\angle B M C$.
Solve as shown in Figure 6, construct a regular $\triangle ABD$ with $AB$ as a side, and connect $CD$.
From the problem, we have $BA = BC$, $\angle ABC = 92^{\circ}$.
Thus, $BA = BC = BD$.
According to Theorem 1, $B$ is the circumcenter of $\angle ACD$.
By Property 2(1), we know
$$
\begin{array}{l}
\angle DCA = \frac{1}{2} \angle DBA = 30^{\circ}, \\
\angle DAC = \frac{1}{2} \angle DBC \\
= \frac{1}{2} (92^{\circ} - 60^{\circ}) = 16^{\circ}.
\end{array}
$$

Therefore, $\angle DCA = \angle MCA$,
$\angle DAC = \angle MAC$.
Thus, $\triangle ADC \cong \triangle AMC$.
Hence, $AM = AD = AB$.
$$
\begin{array}{l}
\text{Then } \angle ABM = \angle AMB \\
= \frac{1}{2} [180^{\circ} - (44^{\circ} - 16^{\circ})] = 76^{\circ}.
\end{array}
$$

Therefore, $\angle BMC = \angle BAC + \angle ABM + \angle MCA$
$$
= 44^{\circ} + 76^{\circ} + 30^{\circ} = 150^{\circ}.
$$