Example 5. On two parallel lines $AB$ and $CD$, take fixed points $M$ and $N$ respectively. On line $AB$, take a fixed-length segment $ME=a$. On the perpendicular segment $MN \perp$, take a point $K$, connect $EK$ and extend it to intersect $CD$ at $F$. If the distance between the two parallel lines is $d$, at what position of $K$ is the sum of the areas of $\triangle MEK$ and $\triangle NFK$ minimized? Also, find this minimum value. (198t, Chongqing Junior High School Competition)
Given the figure, let the length of $MN$ be $l$, the angle between $MN$ and the parallel lines be $\alpha$, and $MK = x$. Then,
$$
\begin{array}{l}
NK = l - x \\
\text{Since } \Delta E M K
\end{array}
$$
$\rightarrow \triangle F N K$,
thus $\frac{a}{F N} = -\frac{x}{l - x}$,
hence $F N = \frac{a(l - x)}{x}$.
$$
\begin{array}{l}
\therefore S_{\triangle \text{ area sum }} = S_{\triangle E M K} + S_{\triangle F N K} \\
= \frac{1}{2} a x \sin \alpha + \frac{1}{2} \cdot \frac{a(l - x)^2}{x} \cdot \sin \alpha \\
= \frac{1}{2} a \sin \alpha \left[ x + \frac{(l - x)^2}{x} \right],
\end{array}
$$

where $\frac{1}{2} a \sin \alpha$ is a constant, and $S$ has the same maximum value as $y = x + \frac{(1 - x)^2}{x}$.
$$
\begin{array}{l}
\because \quad y = x + \frac{(l - x)^2}{x} = 2x - 2l + \frac{l^2}{x} \\
= \left( \sqrt{2x} - \frac{l}{\sqrt{x}} \right)^2 + 2l(\sqrt{2} - 1),
\end{array}
$$

Clearly, when $\sqrt{2} x = \frac{l}{\sqrt{x}} = 0$, the minimum value is $2l(\sqrt{2} - 1)$. From $\sqrt{2x} = -\sqrt{x}$, we get $x = \frac{\sqrt{2}}{2} l$. That is, when $MK = \frac{\sqrt{2}}{2} l$, $S$ has its minimum value,
$$
\begin{aligned}
S_{\text{min}} & = (\sqrt{2} - 1) a l \sin \alpha \\
& = (\sqrt{2} - 1) a d \quad (\text{since } l \sin \alpha = d).
\end{aligned}
$$