Example 5. In the tetrahedron $ABCD$, the plane $ABC$ and the plane $BCD$ are both equilateral triangles with side lengths of $2a$, and $AD=$ $2\sqrt{2}a$. If a point $M$ on $AR$ travels across the surface of the tetrahedron to reach the midpoint $N$ of $CD$, find the shortest path. (1979, Henan Province Mathematics Competition)
Analyzing the four ways to travel from $M$ to $N$ along the surface of a tetrahedron:
(i) When crossing the edge $A C$, unfold $\triangle A B C$ and $\triangle A C D$ into a plane (as shown in the figure). In this case, the line segment $M N$ is the shortest path.

Connecting $C M$, it is clear that $\triangle A C D$ is an isosceles right triangle. Therefore,
$$
\begin{aligned}
\angle M C N & =90^{\circ}+\frac{60^{\circ}}{2}=120^{\circ}, \\
M C & =\sqrt{3} a, C N=a . \text { Thus } \\
M N & =\sqrt{M C^{2}+C N^{2}-2 M C \cdot C N \cos 120^{\circ}} \\
& \approx \sqrt{4+\sqrt{3}} a .
\end{aligned}
$$
(ii) When crossing the edge $B D$, similarly we get
$$
M N=\sqrt{4+\sqrt{3}} a \text {. }
$$
(iii) When crossing the edge $B C$, $M N=A C=2 a$.
(iv) When crossing the edge $A D$, $M N=A C=2 a$.

Therefore, the shortest path length is $2 a$, which goes from $M$ in a straight line to the midpoint of $B C$ (or $A D$), and then in a straight line to $N$.