3. As shown in Figure 1, EF//BC, $S_{\text {EEF }}$ $=S_{\triangle E B C}$. If $S_{\triangle A B C}=1$, then $S_{\triangle C E F}=$ ( ).
(A) $\sqrt{5}-2$
(B) $\sqrt{3}-\frac{3}{2}$
(C) $\frac{1}{4}$
(D) $\frac{1}{5}$
3. (A).

As shown in Figure 6, draw $A Q \perp E F, E P$
$$
\begin{array}{c}
\perp B C \text{. By } A Q \cdot E F=E P \cdot B C \Rightarrow \\
\frac{E F}{B C}=\frac{E P}{A Q}=\frac{E B}{A E} . \\
\text{Also } \frac{E F}{B C}=\frac{A E}{A B} \\
\Rightarrow A E^{2}=A B \cdot E B,
\end{array}
$$
$E$ is the golden section point of $A B$, and $\frac{A E}{A B}=\frac{\sqrt{5}-1}{2}$.
$$
\therefore S_{C E E F}=1-2\left(\frac{\sqrt{5}-1}{2}\right)^{2}=\sqrt{5}-2 \text{. }
$$