For example, the area of $\triangle ABC$ is $1$, points $D$ and $E$ are on sides $AB$ and $AC$ respectively, $BE$ and $CD$ intersect at point $P$, and the area of quadrilateral $BCED$ is twice the area of $\triangle PBC$. Find the maximum value of the area of $\triangle PDE$.
Let $\frac{A D}{A B}=x, \frac{A E}{A C}=y$, then $S_{\triangle A D E}=x y, S_{B C D E}=1-x y$,
$$
\begin{aligned}
S_{\triangle P B C} & =\frac{1}{2}(1-x y), \\
\frac{S_{\triangle P D E}}{S_{P B C}} & =\frac{P E \cdot P D}{P B \cdot P C} \\
& =\frac{S_{\triangle A P E}}{S_{\triangle A B P}} \cdot \frac{S_{\triangle A P D}}{S_{\triangle A P C}} \\
& =\left(\frac{S_{\triangle A P D}}{S_{\triangle A B P}}\right) \cdot\left(\frac{S_{\triangle A P E}}{S_{\triangle A P C}}\right) \\
& =x y .
\end{aligned}
$$

Thus, $S_{\triangle P D E}=\frac{1}{2} x y(1-x y)$.
By Menelaus' theorem, $\frac{B P}{P E} \cdot \frac{E C}{C A} \cdot \frac{A D}{D B}=1$, so
$$
\begin{array}{l}
\frac{B P}{P E}=\frac{1-x}{x(1-y)}, \frac{B P}{B E}=\frac{1-x}{x(1-y)+1-x}=\frac{1-x}{1-x y}, \\
S_{\triangle B P C}=\frac{B P}{B E} S_{\triangle B C E}=\frac{B P}{B E} \times \frac{E C}{A C} S_{\triangle A B C}=\frac{(1-x)(1-y)}{1-x y} .
\end{array}
$$

We now need to find the maximum value of $S=\frac{1}{2} x y(1-x y)$ under the condition $\frac{1}{2}(1-x y)=\frac{(1-x)(1-y)}{1-x y}$.

Let $u=x y$, (since $0<x, y<1$, we have $0<u<1$), then the condition becomes $\frac{1}{2}(1-u)^{2}=1+u-(x+y) \leqslant 1+u-2 \sqrt{u}=(1-\sqrt{u})^{2}$, solving this gives $0<\sqrt{u} \leqslant \sqrt{2}-1$, i.e., $0<u \leqslant 3-2 \sqrt{2}$. The quadratic function $f(u)=\frac{1}{2} u(1-u)$ is increasing on $(0,3-2 \sqrt{2})$, so the maximum value is $f(3-2 \sqrt{2})=5 \sqrt{2}-7$.

Therefore, when $x=y=\sqrt{2}-1$ (at this point $x y=3-2 \sqrt{2}$), the area of $\triangle P D E$ is maximized, with the maximum value being $5 \sqrt{2}-7$.