Example 2: With point $C$ as the center of rotation, rotate $\triangle A B C$ to $\triangle A^{\prime} B^{\prime} C$. Let the midpoints of segments $B A^{\prime}$, $A C$, and $B^{\prime} C$ be $M$, $E$, and $F$ respectively. If $A C \neq B C$ and $E M = F M$, find the degree measure of $\angle E M F$.
(2005, Bulgarian National Mathematical Olympiad)
Proof: As shown in Figure 3, connect \( A A^{\prime} \) and \( B B^{\prime} \). Let
\[
\begin{array}{l}
B C = B^{\prime} C = a, \\
A C = A^{\prime} C = b, \\
A B = A^{\prime} B^{\prime} = c, \\
\angle A C B = \gamma.
\end{array}
\]

Since \(\triangle A A^{\prime} C \sim \triangle B B^{\prime} C\), we have
\[
\begin{array}{l}
A A^{\prime} : A C \\
= B B^{\prime} : B C = k.
\end{array}
\]

Thus, \(A A^{\prime} = k b\) and \(B B^{\prime} = k a\).
Applying Euler's theorem in the quadrilaterals \(A B C A^{\prime}\) and \(A^{\prime} B^{\prime} B C\), we get
\[
\begin{array}{l}
4 E M^{2} = k^{2} b^{2} + c^{2} + b^{2} + a^{2} - A^{\prime} B^{2} - b^{2}, \\
4 F M^{2} = k^{2} a^{2} + c^{2} + b^{2} + a^{2} - A^{\prime} B^{2} - a^{2}.
\end{array}
\]

Since \(E M = F M\), we have
\[
\left(k^{2} - 1\right)\left(a^{2} - b^{2}\right) = 0.
\]

Given that \(a \neq b\), it follows that \(k = 1\).
Therefore, \(\triangle A A^{\prime} C\) is an equilateral triangle, meaning the rotation angle is \(\pm 60^{\circ}\).

We will consider the case where the rotation angle is \(60^{\circ}\) (counterclockwise), the other case is similar.
In \(\triangle A^{\prime} B C\), by the cosine rule, we get
\[
A^{\prime} B^{2} = a^{2} + b^{2} - 2 a b \cos \left(60^{\circ} - \gamma\right).
\]

In \(\triangle A B^{\prime} C\), we have
\[
4 E F^{2} = B^{\prime} A^{2} = a^{2} + b^{2} - 2 a b \cos \left(60^{\circ} + \gamma\right).
\]

Since \(4 E M^{2} = c^{2} + b^{2} + a^{2} - A^{\prime} B^{2}\),
\[
\begin{array}{l}
= c^{2} + 2 a b \cos \left(60^{\circ} - \gamma\right) \\
= a^{2} + b^{2} - 2 a b \cos \gamma + 2 a b \cos \left(60^{\circ} - \gamma\right) \\
= a^{2} + b^{2} - 2 a b \cos \left(60^{\circ} + \gamma\right) = 4 E F^{2},
\end{array}
\]

Thus, \(\triangle E F M\) is an equilateral triangle.
Therefore, \(\angle E M F = 60^{\circ}\).