Initial 74. Given isosceles $\triangle A B C$ with vertex angle $A$ being $108^{\circ}, D$ is a point on the extension of $A C$, and $A D=B C, M$ is the midpoint of $B D$. Find the degree measure of $\angle C M A$.
Given that $AB = AC$, $\angle ABC = \angle ACB = 36^\circ$.

Construct the angle bisector $BP$ of $\angle ABC$ intersecting $AC$ at $P$. Take a point $Q$ on $BC$ such that $BQ = AB$. Connect $PQ$. From the construction, we know that $\triangle ABP \cong \triangle QBP$.
$$
\begin{array}{l}
\text{Therefore, } \angle PQB = \angle PAB = \angle CAB = 108^\circ, \\
\angle PQC = 180^\circ - \angle PQB = 72^\circ. \\
\angle CPQ = 180^\circ - \angle PQC - \angle PCQ = \\
180^\circ - 72^\circ - 36^\circ = 72^\circ.
\end{array}
$$

Thus, $\angle PQC = \angle CPQ$, which means $CP = CQ$.
Hence, $BC = BQ + QC = AB + PC = AC + PC$.
But $BC = AD = AC + CD$.
Therefore, $PC = CD$.
Thus, $C$ is the midpoint of $PD$, so $MC \parallel BP$, and $\angle BCM = \angle PBC = \frac{1}{2} \times 36^\circ = 18^\circ$.
Let $N$ be the midpoint of $BC$. Connect $AN$ and $MN$. It is easy to see that $AN$ is the perpendicular bisector of $BC$ and $MN \parallel AD$.
Thus, $\angle CAN = 54^\circ$, $\angle ANC = 90^\circ$, and
$$
\begin{array}{l}
\angle MNC = \angle ACB = 36^\circ \\
\angle NMC = 180^\circ - \angle CNM - \angle NCM = \\
180^\circ - \angle CNM - \angle BCM = 180^\circ - 36^\circ - 18^\circ
\end{array}
$$
$= 126^\circ$,
Therefore, $\angle CAN + \angle NMC = 180^\circ$.
Hence, points $A$, $N$, $M$, and $C$ are concyclic.
Thus, $\angle CMA = \angle ANC = 90^\circ$.
That is, $\angle CMA = 90^\circ$.
(Zhou Wenshu, Institute of Mathematics, Jilin University, 130023)