10. As shown in Figure 2, the side length of rhombus $A B C D$ is 1, and $\angle A B C=120^{\circ}$. If
$E$ is any point on the extension of $B C$,
$A E$
intersects $C D$ at point $F$, then
the angle between vector $B F$ and $E D$
is
$\qquad$
10.120.

As shown in Figure 5, establish a Cartesian coordinate system. Then
$$
\begin{array}{l}
A\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right), \\
B(0,0), \\
C(1,0), \\
D\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) .
\end{array}
$$

Let $E(a, 0)(a>1)$. Then
$l_{c p}: y=-\sqrt{3}(x-1)$,
$$
l_{\text {AE }}: y=-\frac{\sqrt{3}}{2 a+1}(x-a) \text {. }
$$

By solving the two equations simultaneously, we get the intersection point of line $C D$ and line $A E$
$$
F\left(\frac{a+1}{2 a}, \frac{\sqrt{3}(a-1)}{2 a}\right) \text {. }
$$

Thus, $B F=\left(\frac{a+1}{2 a}, \frac{\sqrt{3}(a-1)}{2 a}\right)$,
$$
E D=\left(\frac{1-2 a}{2}, \frac{\sqrt{3}}{2}\right) \text {. }
$$
$B F \cdot E D=-\frac{a^{2}-a+1}{2 a},|B F|=\frac{\sqrt{a^{2}-a+1}}{a}$,
$$
|E D|=\sqrt{a^{2}-a+1} \text {. }
$$

Let $\theta$ be the angle between $\boldsymbol{B F}$ and $\boldsymbol{E D}$, then $\cos \theta=\frac{\boldsymbol{B F} \cdot \boldsymbol{E} \boldsymbol{D}}{|\boldsymbol{B F}| \cdot|E D|}=-\frac{1}{2}$, which means $\theta=120^{\circ}$.