Problem 365

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 $x=\sqrt{5}-2$.