10. 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 $\angle A B F=$
$90^{\circ}$
10.【Analysis and Solution】Being sensitive to data, one would notice that $\frac{c}{a}=\frac{\sqrt{5}-1}{2}=\frac{-1+\sqrt{1^{2}-4 \times 1 \times(-1)}}{2 \times 1}$ is a root of the equation $x^{2} + x - 1 = 0$. Substituting and simplifying, we get $c^{2} + a c - a^{2} = 0$, which leads to $a c = b^{2}$. This precisely matches the projection theorem, thus $\angle A B F = 90^{\circ}$.