Example 10 In $\triangle A B C$ with area 1, points $D, E$ are on sides $A B, A C$ respectively, and line segments $B E, C D$ intersect at point $P$. Points $D, E$ move on $A B, A C$ respectively, satisfying the condition that the area of quadrilateral $B C E D$ is twice the area of $\triangle P B C$. Find the maximum value of the area of $\triangle P D E$.

In $\triangle A B C$ with area 1, points $D, E$ are on sides $A B, A C$ respectively, and line segments $B E, C D$ intersect at point $P$. Points $D, E$ move on $A B, A C$ respectively, satisfying the condition that the area of quadrilateral $B C E D$ is twice the area of $\triangle P B C$. Find the maximum value of the area of $\triangle P D E$.
Consider $\triangle PDE, \triangle PBC, \triangle PBD, \triangle PCE$ with areas $S_{1}, S_{2}, S_{3}, S_{4}$ respectively. Let $AD: AB = \lambda_{1}$ and $AE: AC = \lambda_{2}$. Then, $S_{\triangle ADE}: S_{\triangle ABC} = \lambda_{1} \lambda_{2}$, which means
$$
S_{\triangle ADE} = \lambda_{1} \lambda_{2}, \quad S_{2} = \frac{1}{2}\left(1 - \lambda_{1} \lambda_{2}\right)
$$

Noting the conclusion 1 from the first chapter of the first section, i.e., $S_{\triangle ADE}: S_{\triangle ABC} = S_{1}: S_{2}$, we get
$$
S_{1} = S_{2} \cdot S_{\triangle ADE} = \frac{1}{2}\left(1 - \lambda_{1} \lambda_{2}\right) \cdot \lambda_{1} \lambda_{2} = -\frac{1}{2}\left(\lambda_{1} \lambda_{2} - \frac{1}{2}\right)^{2} + \frac{1}{8}
$$

Also, $S_{1}: S_{3} = PE: PB = S_{4}: S_{2}$, which implies $S_{1} \cdot S_{2} = S_{3} \cdot S_{4}$. Since $S_{2} = S_{1} + S_{3} + S_{4}$, we have

Thus,
$$
\begin{array}{c}
S_{2} - S_{1} = S_{3} + S_{4} \geqslant 2 \sqrt{S_{3} S_{4}} = 2 \sqrt{S_{1} S_{2}} \\
S_{1} - 6 S_{1} S_{2} + S_{2}^{2} \geqslant 0 \Rightarrow \left(\lambda_{1} \lambda_{2}\right)^{2} - 6 \lambda_{1} \lambda_{2} + 1 \geqslant 0
\end{array}
$$

Since $0 < \lambda_{1} \lambda_{2} < 1$, we have $\lambda_{1} \lambda_{2} \leqslant 3 - 2 \sqrt{2} \in \left(-\infty, \frac{1}{2}\right)$. Therefore, when $\lambda_{1} \lambda_{2} = 3 - 2 \sqrt{2}$, $S_{1}$ achieves its maximum value, which is $5 \sqrt{2} - 7$.