Example 4 As shown in Figure 2, there is a rectangular plot of land $A B C D$, and a rectangular flower bed $E F G H$ is to be built in the center, with its area being half of the area of this plot, and the width of the paths around the flower bed being equal. Now, without any measuring tools, only a sufficiently long rope with equal markings, how can the width of the path be measured?
Solution: Let the width of the road be $x$, $AB=a$, $AD=b$. According to the problem, we have
$$
(a-2 x)(b-2 x)=\frac{1}{2} a b \text{. }
$$

Thus, $8 x^{2}-4(a+b) x+a b=0$.
Solving for $x$ gives $x=\frac{1}{4}\left[(a+b) \pm \sqrt{a^{2}+b^{2}}\right]$.
When the plus sign is taken,
$$
\begin{aligned}
2 x & =\frac{1}{2}\left[(a+b)+\sqrt{a^{2}+b^{2}}\right] \\
& >\frac{(a+b)+b}{2}>b, \text{ which is not feasible. }
\end{aligned}
$$

Therefore, we take $x=\frac{1}{4}\left[(a+b)-\sqrt{a^{2}+b^{2}}\right]$.
From the width $x=\frac{1}{4}\left[(a+b)-\sqrt{a^{2}+b^{2}}\right]$, we get the following method:

Measure the length of $AB^* + AD$ (i.e., $a+b$) with a rope, subtract the length of $BD$ (the diagonal $BD=\sqrt{a^{2}+b^{2}}$), to get $l=AB+AD-BD$. Then fold $l$ in half twice, which gives the width of the road.
$$
\begin{array}{l}
\text{Thus, } x=\frac{AB+AD-BD}{4} \\
= \frac{a+b-\sqrt{a^{2}+b^{2}}}{4} .
\end{array}
$$