Example 2 As shown in Figure 3, in rectangle $A B C D$, $A B=20$, $B C=10$. 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.
Solve as shown in Figure 4, construct the symmetric point $B'$ of point $B$ with respect to $AC$, and connect $AB'$. Construct $B'N \perp AB$, intersecting $AC$ at point $M$, and connect $BM$. Then $BM=B'M$.
Therefore, $BM+MN=B'M+MN$.
By the shortest distance of a perpendicular segment, we know that $B'M+MN$ is minimized, so $BM+MN$ also achieves its minimum value.
It is easy to see that $\angle BAC=\angle PAC, \angle BAC=\angle ACP$.
Thus, $\angle PAC=\angle ACP$. Therefore, $PA=PC$.
Let $PA=PC=x$. Then $PD=20-x$.
In the right triangle $\triangle ADP$, by $AD^2+PD^2=AP^2$, we get $10^2+(20-x)^2=x^2$.
Solving for $x$ yields $x=12.5$.
Thus, $PD=7.5$.
It is easy to see that $\triangle ADP \sim \triangle B'AN$.
By $\frac{B'N}{AD}=\frac{AN}{PD}=\frac{AB'}{AP}=\frac{AB}{AP}$, we get $B'N=16$, which is the minimum value of $BM+MN$. At this time, $AN=12$.