Example 3. There is a convex pentagon, the area of the triangles formed by any three consecutive vertices is 1. Find the area of this pentagon (1972, American Mathematics Competition)
Analyze the figure. Since the area of the triangle formed by any three adjacent vertices is 1, according to the theorem of equal areas of similar triangles, we have $A E / / B D, A D / / B C, B E / / C D, \therefore$ quadrilateral $B C D O$ is a parallelogram. Using the area relationships of parallelograms and trapezoids, the key to finding the area of pentagon $A B C D E$ lies in finding the area of $\triangle A O E$. Therefore, solving this problem requires finding the equation model for $S_{\triangle A O E}$.

Solution: $\because$ Quadrilateral $B C D O$ is a parallelogram,
$$
\therefore S_{\triangle B O D}=1.
$$

Let $S_{\triangle A E O}=x$, then the area of trapezoid $A B D E$ is $3-x$. In trapezoid $A B D E$, $S_{\triangle A B O}=S_{\triangle D E O},\left(S_{\triangle A B O}\right)^{2}=\frac{1}{2} O A \cdot O B \sin \angle A O B \cdot \frac{1}{2} O D \cdot D E \sin \angle D O E=\frac{1}{2} O A \cdot OE \sin \angle A O E \cdot \frac{1}{2} O B \cdot O D \sin \angle B O D$ $=S_{\triangle B O D} \cdot S_{\triangle A O E}=1 \cdot x$,
$$
\begin{array}{l}
\therefore S_{\triangle A B O}=S_{\triangle D E O}=\sqrt{1 \cdot x} ; \\
S_{\triangle B D E}=S_{\triangle B O D}+S_{\triangle A O E}+2 S_{\triangle A B D} \\
=S_{\triangle B O D}+S_{\triangle A O E}+2 \sqrt{S_{\triangle B O D} \cdot S_{\triangle A O E}} \\
=\left(S_{\triangle B O D}+S_{\triangle A O E}\right)^{2} . \\
\therefore(\sqrt{x}+1)^{2}=3-x .
\end{array}
$$

Solving for $\sqrt{x}=\frac{\sqrt{5}-1}{2}$, thus the area of the convex pentagon is
$$
\begin{aligned}
S & =S_{A B D E}+S_{\triangle B C D} \\
& =\frac{3+\sqrt{5}}{2}+1=\frac{5+\sqrt{5}}{2} .
\end{aligned}
$$