2. We call an ellipse with an eccentricity equal to the golden ratio $\frac{\sqrt{5}}{2}-1$ a "beautiful ellipse". Let $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>i$ $>0$ ) be a beautiful ellipse, $F, A$ are its left focus and vertex, respectively, and $B$ is one endpoint of its minor axis. Then $\angle A B F$ equals ( ).
(A) $60^{\circ}$
(B) $75^{\circ}$
(C) $90^{\circ}$
(D) $120^{\circ}$
2. (C).

From $\frac{c}{a}=\frac{\sqrt{5}-1}{2}$, we get $a^{2}-a c-c^{2}=0$.
$$
\begin{aligned}
\because \quad & |A B|^{2}+|B F|^{2}-|F A|^{2} \\
& =\left(a^{2}+b^{2}\right)+a^{2}-(a+c)^{2} \\
& =2\left(a^{2}-a c-c^{2}\right)=0, \\
\therefore \quad & |A B|^{2}+|B F|^{2}=|F A|^{2}, \text{ hence } \angle A B F=90^{\circ} .
\end{aligned}
$$