Example 7 Two residential areas $A$ and $B$ on the same side of the riverbank $l$ are at distances of $a \, \text{m}$ and $b \, \text{m}$ from the riverbank (i.e., in Figure 7, $A A^{\prime}=a \, \text{m}, B B^{\prime}=b \, \text{m}$), and $A^{\prime} B^{\prime}=c \, \text{m}$. A green belt $C D$ of length $s \, \text{m}$ (negligible width) is to be built on the riverbank, such that the sum of the distances from $C$ to area $A$ and from $D$ to area $B$ is minimized.
(1) Draw the position of the green belt in Figure 8, and describe the drawing process;
(2) Find the minimum value of $A C + B D$.
(20th Jiangsu Province Junior High School Mathematics Competition)
(1) As shown in Figure 9, draw a line segment $A P / / l$, such that $A P = s$. Take the symmetric point of $P$ with respect to $l$ as $P^{\prime}$, and connect $B P^{\prime}$, intersecting $l$ at point $D$. On $l$, to the left of point $D$, intercept $D C = s$. Then $C D$ is the position of the required green belt.

In fact, as shown in Figure 9, assume the green belt is built at another position $C^{\prime} D^{\prime}$. Connect $B D^{\prime}$, $P D^{\prime}$, $A C^{\prime}$, and $P^{\prime} D^{\prime}$.
Then, by symmetry, we have
$$
P^{\prime} D = P D, \quad P^{\prime} D^{\prime} = P D^{\prime}.
$$

Since $A P \perp C D$ and $A P \perp C^{\prime} D^{\prime}$, we know $A C = P D$ and $A C^{\prime} = P D^{\prime}$.
But $P^{\prime} D^{\prime} + B D^{\prime} \geqslant P^{\prime} B = P^{\prime} D + B D$
$\Rightarrow P^{\prime} D + B D^{\prime} \geqslant P D + B D$
$$
\Rightarrow A C + B D \leqslant A C^{\prime} + B D^{\prime}.
$$
(2) Draw $P^{\prime} H \perp B B^{\prime}$ at point $H$. Then
$$
B H = B B^{\prime} + B^{\prime} H = b + a, \quad P^{\prime} H = c - s.
$$

Thus, $P^{\prime} B = \sqrt{P^{\prime} H^{2} + B H^{2}}$
$$
= \sqrt{(c-s)^{2} + (b+a)^{2}}.
$$

Therefore, the minimum value of $A C + B D$ is
$$
\sqrt{(c-s)^{2} + (b+a)^{2}}.
$$