8. In $\triangle ABC$ with area 1, points $D$ and $E$ are on sides $AB$ and $AC$ respectively. Segments $BE$ and $CD$ intersect at point $P$. Points $D$ and $E$ move on $AB$ and $AC$ respectively, but satisfy the condition that the area of quadrilateral $BCED$ is twice the area of $\triangle PBC$. Find the maximum area of $\triangle PDE$.
(1992 Japan Mathematical Olympiad)
8. Let $\frac{A D}{D B}=\frac{s}{1-s}, \frac{A E}{E C}=\frac{t}{1-t}(0<s<1,0<t<1)$.

For $\triangle C D A$ and line $B P E$, applying Menelaus' theorem, we have $\frac{C P}{P D} \cdot \frac{D B}{B A} \cdot \frac{A E}{E C}=1$, which simplifies to $\frac{C P}{P D}=\frac{1-t}{t(1-s)}$.
For $\triangle B E A$ and line $C P B$, applying Menelaus' theorem, we have $\frac{B P}{P E} \cdot \frac{E C}{C A} \cdot \frac{A D}{D B}=1$, which simplifies to $\frac{B P}{P E}=\frac{1-s}{s(1-t)}$.
And $S_{B C E D}=S_{\triangle A B C}-S_{\triangle A D E}=\frac{1}{2} A B \cdot A C \cdot \sin A-\frac{1}{2} A D \cdot A E \cdot \sin A=(1-s t) \cdot \frac{1}{2} A B \cdot A C \cdot \sin A=(1-s t)$. $S_{\triangle A B C}=1-s t, S_{\triangle P B C}=\frac{1-t}{(1-t)+t(1-s)} \cdot S_{\triangle B D C}=\frac{1-t}{1-s t} \cdot(1-s) \cdot S_{\triangle A B C}=\frac{(1-s)(1-t)}{1-s t}$, given that $S_{B C E D}=2 S_{\triangle P B C}$, we get $1-s t=2 \cdot \frac{(1-s)(1-t)}{1-s t}$, thus $(1-s t)^{2}=2(1-s)(1-t)=2[1-(s+t)+s t]$. Also, $S_{\triangle P D E}=\frac{t \cdot(1-s) \cdot s \cdot(1-t)}{(1-s)(1-t)} \cdot S_{\triangle P B C}=s t \cdot S_{\triangle P B C}=s t \cdot \frac{1}{2} S_{B C E D}=\frac{1}{2} s t(1-s t)$.
This problem requires finding the maximum value of (2) under condition (1).
Using the AM-GM inequality $s+t \geqslant 2 \sqrt{s t}$ on (1), we have
$$
(1-s t)^{2}=2[1-(s+t)+s t] \leqslant 2(1-2 \sqrt{s t}+s t)=2(1-\sqrt{s t})^{2} \text {, }
$$
$\therefore \quad 1-s t \leqslant \sqrt{2}(1-\sqrt{s t})$, which leads to $(\sqrt{s t}-1) \cdot[\sqrt{s t}-(\sqrt{2}-1)] \geqslant 0$.
$\because 0<s, t<1$, then $\sqrt{s t}-1<0$, thus $0<\sqrt{s t} \leqslant \sqrt{2}-1$. Equality holds when $s=t=\sqrt{2}-1$.
$S_{\triangle P D E}=\frac{1}{2} s t(1-s t)=-\frac{1}{2}\left(s t-\frac{1}{2}\right)^{2}+\frac{1}{8}$.
When $s t=(\sqrt{2}-1)^{2}=3-2 \sqrt{2}$, $S_{\triangle P D E}$ reaches its maximum value, which is $-\frac{1}{2}\left(3-2 \sqrt{2}-\frac{1}{2}\right)^{2}+\frac{1}{8}=5 \sqrt{2}-7$.