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

Let $S_{\triangle C E F}=x$.
Then $S_{\triangle A E F}=S_{\triangle B C E}=\frac{1-x}{2}, S_{\triangle A B C}=\frac{1+x}{2}$.
Thus, $\frac{A F}{A C}=\frac{S_{\triangle A E F}}{S_{\triangle M B C}}=\frac{1-x}{1+x}$.
But $E F \parallel B C,\left(\frac{A F}{A C}\right)^{2}=\frac{S_{\triangle A E F}}{S_{\triangle A B C}}=\left(\frac{1-x}{1+x}\right)^{2}$,
and $\frac{S_{\triangle A E F}}{S_{\triangle A B C}}=\frac{1-x}{2}$, solving for $x$ yields $x=\sqrt{5}-2$.