Example 10 In trapezoid $ABCD$  
the lower base $AB$ has two fixed points $M$ and $N$, and the upper base $CD$ has a moving point $P$. Let $E = DN \cap AP, F = DN \cap MC, G = MC \cap PB, DP = \lambda DC$. For what value of $\lambda$ is the area of quadrilateral $PEFG$ maximized?
Solution: Let $DC=1$, then $DP=\lambda, PC=1-\lambda$. Let $AM=a, MN=b, NB=c$, and the height of the trapezoid be $h$. Draw $CQ \parallel PB$, intersecting the extension of $AB$ at $Q$ (as shown in Figure 2112). $PCQB$ is a parallelogram, so $BQ=PC=1-\lambda$. Therefore, 
\[
\frac{S_{\triangle MGB}}{S_{MCQ}} = \frac{(b+c)^2}{(b+c+1-\lambda)^2}.
\]
Since $S_{\triangle MCQ} = \frac{1}{2} h(b+c+1-\lambda)$, we have 
\[
S_{\triangle MGB} = \frac{(b+c)^2}{2(b+c+1-\lambda)}.
\]
Similarly, we get 
\[
S_{\triangle AEN} = \frac{(a+b)^2 h}{2(a+b+\lambda)}.
\]
Since 
\[
S_{PEFG} = S_{\triangle APB} - (S_{\triangle MGB} + S_{\triangle AEN}) + S_{\triangle FMN},
\]
and $S_{\triangle APB}$ and $S_{\triangle FMN}$ are constants, $S_{PEFG}$ reaches its maximum value when $S_{\triangle MGB} + S_{\triangle AEN}$ is minimized. We have 
\[
S_{\triangle MGB} + S_{\triangle AEN} = \frac{h}{2} \left[ \frac{(a+b)^2}{a+b+\lambda} + \frac{(b+c)^2}{b+c+1-\lambda} \right].
\]
Using the Cauchy-Schwarz inequality, we get
\[
\begin{array}{l}
\left[ \frac{(a+b)^2}{a+b+\lambda} + \frac{(b+c)^2}{b+c+1-\lambda} \right] [(a+b+\lambda) + (b+c+1-\lambda)] \\
\geqslant [(a+b) + (b+c)]^2 = (a+2b+c)^2,
\end{array}
\]
where equality holds if and only if 
\[
\frac{a+b}{a+b+\lambda} = \frac{b+c}{b+c+1-\lambda},
\]
i.e., 
\[
\lambda = \frac{a+b}{a+2b+c} = \frac{AN}{AN+MB}.
\]
Thus, 
\[
S_{\triangle MGB} + S_{\triangle AEN} \geqslant \frac{h}{2} \frac{(a+2b+c)^2}{a+2b+c+1},
\]
and $S_{\triangle MGB} + S_{\triangle AEN}$ reaches its minimum value when 
\[
\lambda = \frac{AN}{AN+MB}.
\]
At this point, $S_{PEFG}$ reaches its maximum value.

In conclusion, when and only when 
\[
\lambda = \frac{AN}{AN+MB},
\]
the area of quadrilateral $PEFG$ is maximized.