14. As shown in the figure, in the convex pentagon $A B C D E$, the areas of $\triangle A B C$, $\triangle B C D$, $\triangle C D E$, $\triangle D E A$, and $\triangle E A B$ are all equal to 1. Then the area of pentagon $A B C D E$ is $\qquad$.
Answer $\frac{1}{2}(5+\sqrt{5})$.
Analysis As shown in the analysis diagram, let the intersection of $B D$ and $C E$ be $P$. Since the area of $\triangle A B C$ = the area of $\triangle E A B$ = 1, $A B \| E C$. Similarly, it can be proven that $B D \| A E$, so $A B P E$ is a parallelogram, and thus the area of $\triangle B P E$ = the area of $\triangle E A B$ = 1.
Let the area of $\triangle B C P$ = $x$, then the area of $\triangle E P D$ = $x$, and the area of $\triangle C D P$ = $1-x$.
Since
$$
\frac{\text{the area of } \triangle B C P}{\text{the area of } \triangle C D P} = \frac{B P}{P D} = \frac{\text{the area of } \triangle B P E}{\text{the area of } \triangle P D E}
$$

we have $\frac{x}{1-x} = \frac{1}{x}$, which simplifies to $x^2 = 1 - x$, or $x^2 + x - 1 = 0$. Solving this, we get $x = \frac{\sqrt{5} - 1}{2}$ (the negative root is discarded), so,
the area of pentagon $A B C D E$ = the area of $\triangle C D E$ + 2 × the area of $\triangle E A B$ + the area of $\triangle B C P$
$$
= 3 + \frac{\sqrt{5} - 1}{2} = \frac{1}{2}(5 + \sqrt{5})
$$