69. The area of $\triangle A B C$ is $1, D, E$ are points on sides $A B, A C$ respectively, $B E, C D$ intersect at point $P$, and 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 $\triangle P D E$.

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The area of $\triangle A B C$ is $1, D, E$ are points on sides $A B, A C$ respectively, $B E, C D$ intersect at point $P$, and 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 $\triangle P D E$.
69.
$$\begin{array}{l}
\text { Let } \frac{A D}{A B}=x, \frac{A E}{A C}=y, \text { so } S_{\triangle A D E}=x y ; S_{B C E D}=1-x y ; S_{\triangle P B C}=\frac{1}{2}(1-x y) \\
\frac{S_{\triangle P D E}}{S_{\triangle P B C}}=\frac{P E \cdot P D}{P B \cdot P C}=\left(\frac{P E}{P B}\right) \cdot\left(\frac{P D}{P C}\right)=\frac{S_{\triangle A P E}}{S_{\triangle A B P}} \cdot \frac{S_{\triangle A P D}}{S_{\triangle A P C}}=\frac{S_{\triangle A P E}}{S_{\triangle A P C}} \cdot \frac{S_{\triangle A P D}}{S_{\triangle A B P}}=x y
\end{array}$$

Therefore,
$$S_{\triangle P D E}=\frac{1}{2} x y(1-x y)$$

By Menelaus' theorem,
$$\frac{B P}{P E} \cdot \frac{E C}{C A} \cdot \frac{A D}{D B}=1$$

So
$$\begin{array}{c}
\frac{B P}{P E}=\frac{1-x}{x(1-y)} \\
\frac{B P}{B E}=\frac{1-x}{x(1-y)+1-x}=\frac{1-x}{1-x y} \\
S_{\triangle B P C}=\frac{B P}{B E} S_{\triangle B C E}=\frac{B P}{B E} \cdot \frac{E C}{A C} S_{\triangle A B C}=\frac{(1-x)(1-y)}{1-x y}
\end{array}$$

Thus, the problem is reduced to finding the maximum value of equation (1) under the condition
$$\frac{1}{2}(1-x y)=\frac{(1-x)(1-y)}{1-x y}$$

Let $u=x y$, (since $0<x, y<1$, so $0<x y<1$ ),
equation (2) becomes
$$\frac{1}{2}(1-u)^{2}=1+u-(x+y) \leqslant 1+u-2 \sqrt{x y}=(1-\sqrt{u})^{2}$$

So
$$1-u \leqslant \sqrt{2}(1-\sqrt{u})$$

That is
$$1+\sqrt{u} \leqslant \sqrt{2}$$

So
$$0<\sqrt{u} \leqslant \sqrt{2}-1,0<u \leqslant(\sqrt{2}-1)^{2}=3-2 \sqrt{2}$$

The quadratic function $f(u)=\frac{1}{2} u(1-u)$ achieves its maximum value in the interval $(0,3-2 \sqrt{2}]$ at $f(3-$ $2 \sqrt{2})=5 \sqrt{2}-7$.

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