Example 4. Given $\triangle A B C$ and a line $D E$ parallel to $B C$ intersecting at points $B D E$, the area of $\triangle B D E$ equals $k^{2}$. Then, what is the relationship between $k^{2}$ and the area $S$ of $\triangle A B C$ for the problem to have a solution? How many solutions are there?
(1987, Shanghai)
Solve as shown in Figure 4, let $AD: AB = x$.
Given $DE \| BC$, we know $\triangle ADE \sim \triangle ABC$.
Then $\frac{S_{\triangle DRE}}{S_{\triangle ABC}} = \frac{AD^2}{AB^2} = x^2$,
Thus $S_{\triangle DE} = x^2 \cdot S$.
Also, $\frac{S_{\triangle ABE}}{S_{\triangle ADB}} = \frac{AB}{AD} = \frac{1}{x}$,
Therefore, $S_{\triangle ABE} = x \cdot S$.

We get $xS - x^2S = k^2$,
Thus $Sx^2 - Sx + k^2 = 0$.
If this equation has a solution, then
$\Delta = (-S)^2 - 4Sk^2 \geq 0$,
We get $k^2 \leq \frac{S}{4}$.

Therefore, when $k^2 = \frac{S}{4}$, the original problem has one solution (at this time $x = \frac{1}{2}$, meaning $DE$ is the midline of $\triangle ABC$), and when $k^2 < \frac{S}{4}$, the original problem has two solutions.