Example 6 In rectangle $A B C D$, $A B=20 \mathrm{~cm}, B C=$ $10 \mathrm{~cm}$. 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.
Solution: As shown in Figure 9, construct the symmetric point $B^{\prime}$ of point $B$ with respect to $A C$, and connect $A B^{\prime}$. Then the symmetric point of point $N$ with respect to $A C$ is point $N^{\prime}$ on $A B^{\prime}$. At this time, the minimum value of $B M + M N$ is equal to the minimum value of $B M + M N^{\prime}$, which is the distance from $B$ to $A B^{\prime}$, $B H^{\prime}$. Therefore, the minimum value of $B M + M N$ is $B H^{\prime}$.
Next, we find the length of $B H^{\prime}$.
Let $A B^{\prime}$ intersect $D C$ at point $P$, and connect $B P$, then
$S_{\triangle A B P} = \frac{1}{2} \times 20 \times 10 = 100$.
Note that $A P = P C$ (think about why), let $A P = x$, then $P C = x, D P = 20 - x$.
According to the Pythagorean theorem, $P A^{2} = D P^{2} + D A^{2}$, i.e., $x^{2} = (20 - x)^{2} + 10^{2}$.
Solving this, we get $x = 12.5$.
Therefore, $B H^{\prime} = \frac{100 \times 2}{12.5} = 16 (\text{cm})$.
Thus, the minimum value of $B M + M N$ is $16 \text{ cm}$.