Four. (20 points) As shown in the figure, in the convex pentagon $A B C D E$, it is known that $S_{\triangle A B C}=1$, and $E C$ // $A B, A D$ // $B C, B E$ // $C D, C A$ // $D E, D B$ // $E A$. Try to find: the area of pentagon $A B C D E$.

---

The translation preserves the original text's line breaks and formatting.
Given, we have
\[
\begin{aligned}
S_{\triangle A C D} & =S_{\triangle C D E}=S_{\triangle D E A}=S_{\triangle E A B}=S_{\triangle A C B}=S_{\triangle A C F} \\
& =1 .
\end{aligned}
\]

Let \( S_{\triangle A E P} = x \). Then \( S_{\triangle D E} = 1 - x \).
Since \( A A E F \) and \( \triangle D E C \) have equal heights on \( A F \) and \( D F \), we have \( \frac{D F}{A F} = \frac{1 - x}{x} \).

Also, \( \triangle D E F \sim \triangle A C F \), so
\[
\frac{S_{\triangle D E F}}{S_{\triangle A C F}} = \left( \frac{D F}{A F} \right)^2 = \frac{(1 - x)^2}{x^2} = 1 - x.
\]

Solving this, we get \( x = \frac{\sqrt{5} - 1}{2} \).
\[
\text{Therefore, } S_{\triangle B C D E} = 3 S_{\triangle A B C} + S_{\triangle A E F} = \frac{5 + \sqrt{5}}{2}.
\]