Example 20 (1998 Beijing Middle School Mathematics Competition Final Question) As shown in Figure 5-20, in rectangle $ABCD$, $AB=20 \text{ cm}, BC=10 \text{ cm}$. Take a point $M$ on $AC$ and a point $N$ on $AB$ such that the value of $BM+MN$ is minimized. Find this minimum value.
Given that this problem involves two moving points $M$ and $N$, we first assume $N$ to be a fixed point. Thus, the problem is transformed into finding a point on $AC$ such that the sum of its distances to two fixed points $B$ and $N$ on the same side is minimized. Using the concept of reflection, we construct the reflection of point $B$ about $AC$, denoted as $B'$. The line $BB'$ intersects $AC$ at $E$. We then connect $B'N$ and let it intersect $AC$ at $M_0$. We connect $M_0B$ and $MB'$. It is easy to prove that $MB + MN = MB' + MN \geq B'N$. Therefore, to minimize $BM + MN$, we need to minimize $B'N$. We draw a perpendicular line from $B'$ to $AB$, intersecting $A$ at $M'$, with the foot of the perpendicular being $N'$. It is evident that $B'$ is a fixed point and $N$ is a moving point on $AB$. By the property that the shortest distance from a point to a line is the perpendicular distance, we have $B'N \geq B'N'$. Hence, when $M$ and $N$ are at the positions of $M'$ and $N'$ respectively, the value of $BM + MN$ is minimized. It is easy to find that $BE = 4\sqrt{5}$, and $\triangle B'N'B \sim \triangle ABC$. Therefore, we can conclude that $B'N = 16$, so the minimum value of $BM + MN$ is 16.