Example 4 (2000 National High School Competition Question) In the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, let the left focus be $F$, the right vertex be $A$, and the upper endpoint of the minor axis be $B$. If the eccentricity of the ellipse is $\frac{\sqrt{5}-1}{2}$, then $\angle A B F=$
Solution: Fill in $90^{\circ}$. Reason: Let $c$ be the semi-focal distance of the ellipse, then the coordinates of points $A$, $B$, and $F$ are $A(a, 0)$, $B(0, b)$, and $F(-c, 0)$.
From $\frac{c}{a}=\frac{\sqrt{5}-1}{2}$, we get $c^{2}+a c=a^{2}$.
Since $\overrightarrow{B F} \cdot \overrightarrow{B A}=(-c,-b) \cdot(a,-b)=-a c+b^{2}=-a c+a^{2}-c^{2}=0$,
we know $\overrightarrow{B F} \perp \overrightarrow{B A}$. Therefore, $\angle A B F=90^{\circ}$.