Example 9 There are 4 factories $A, B, C, D$, and $AB = a \text{ km}, BC = \frac{\sqrt{2}}{2} a \text{ km}, CD = \frac{\sqrt{2}}{4} a \text{ km}, \angle ACB = 90^\circ, \angle BCD = 120^\circ$. Now we need to find a location for a supply station $H$ such that the sum of its distances to the 4 factories, $HA + HB + HC + HD$, is minimized. Explain the reasoning and find the minimum value.
(2000, Huanggang City, Hubei Province, Junior High School Mathematics Competition)
(1) As shown in Figure 6, when $A$ and $D$ are on opposite sides of line $BC$, take any point $H'$ in $\triangle ABD$ other than $C$, and connect $H'A$, $H'B$, $H'C$, and $H'D$. Then,
$$
\begin{array}{l}
H'B + H'C > BC, \\
H'A + H'D > CA + CD.
\end{array}
$$

Thus, $H'B + H'C + H'A + H'D$
$$
> CA + CB + CD.
$$

Therefore, when $H$ is at point $C$, $HA + HB + HC + HD$ is minimized, and the minimum value is
$$
\begin{array}{l}
CA + CB + CD \\
= \sqrt{a^2 - \left(\frac{\sqrt{2}}{2} a\right)^2} + \frac{\sqrt{2}}{2} a + \frac{\sqrt{2}}{4} a = \frac{5 \sqrt{2}}{4} a.
\end{array}
$$

(2) As shown in Figure 7, when $A$ and $D$ are on the same side of $BC$, take any point $H'$ in quadrilateral $ABCD$ other than the intersection of the diagonals, and connect $H'A$, $H'C$, $H'B$, and $H'D$. Then,
$$
\begin{array}{l}
H'A + H'C > AC, \\
H'B + H'D > BD.
\end{array}
$$

Thus, $H'A + H'B + H'C + H'D > AC + BD$. Therefore, when $H$ is at the intersection of $AC$ and $BD$,
$$
HA + HB + HC + HD = AC + BD
$$

is the minimum value.
Draw $DM \perp BC$ at $M$. In right $\triangle DCM$,
$$
\begin{array}{l}
DC = \frac{\sqrt{2}}{4} a, \angle DCM = 60^\circ, \\
CM = DC \cdot \cos 60^\circ = \frac{\sqrt{2}}{8} a, \\
DM = DC \cdot \sin 60^\circ = \frac{\sqrt{6}}{8} a.
\end{array}
$$

Then, $BD = \sqrt{BM^2 + DM^2}$
$$
= \sqrt{\left(\frac{\sqrt{2}}{2} a + \frac{\sqrt{2}}{8} a\right)^2 + \left(\frac{\sqrt{6}}{8} a\right)^2} = \frac{\sqrt{14}}{4} a.
$$

At this point, the minimum value of $HA + HB + HC + HD$ is
$$
AC + BD = \frac{2 \sqrt{2} + \sqrt{14}}{4} a.
$$