Example 5 As shown in Figure 4, in rectangle $A B C D$, $A B=$ $20 \text{ cm}, B C=10 \text{ cm}$. If points $M$ and $N$ are taken on $A C$ and $A B$ respectively, such that the value of $B M+M N$ is minimized, find this minimum value.
(1998, Beijing Junior High School Mathematics Competition)
Solution: As shown in Figure 4, construct $\angle B_{1} A C = \angle B A C$, draw $B Q \perp A C$ at $Q$, and extend $B Q$ to intersect $A B_{1}$ at $B_{1}$; draw $N P \perp A C$ at $P$, and extend $N P$ to intersect $A B_{1}$ at $N_{1}$, then points $B_{1}$ and $B$, and points $N_{1}$ and $N$ are symmetric with respect to $A C$. Draw $B H \perp A B_{1}$ at $H$.
It is easy to see that $\triangle M P N \cong \triangle M P N_{1}$.
Thus, $M N_{1} = M N$.
Therefore, $B M + M N = B M + M N_{1}$
$$
\geqslant B N_{1} \geqslant B H \text {. }
$$

Hence, the minimum value of $B M + M N$ is the distance from point $B$ to $A B_{1}$.

In the right triangle $\triangle A B C$,
$$
B Q = \frac{A B \cdot B C}{A C} = 4 \sqrt{5}, B B_{1} = 2 B Q = 8 \sqrt{5} \text {. }
$$

In the right triangle $\triangle A B Q$,
$$
A Q = \sqrt{A B^{2} - B Q^{2}} = 8 \sqrt{5} \text {. }
$$

In $\triangle A B B_{1}$,
$$
B H = \frac{A Q \cdot B B_{1}}{A B_{1}} = \frac{8 \sqrt{5} \times 8 \sqrt{5}}{20} = 16 \text {. }
$$

Therefore, the minimum value of $B M + M N$ is $16 \text{ cm}$.