9. As shown in Figure 4, rays $A M$ and $B N$ are both perpendicular to line segment $A B, E$ is a point on $A M$, and a perpendicular line $A C$ is drawn from point $A$ to $B E$, intersecting $B E$ and $B N$ at points $F$ and $C$, respectively. A perpendicular line $C D$ is drawn from point $C$ to $A M$, with the foot of the perpendicular being $D$. If $C D = C F$, then the value of $\frac{A E}{A D}$ is $\qquad$.
9. $\frac{\sqrt{5}-1}{2}$.

Let $F C=m, A F=n$.
Since Rt $\triangle A F B \backsim$ Rt $\triangle A B C$, we have $A B^{2}=A F \cdot A C$.
Also, since $F C=D C=A B$, we have $m^{2}=n(n+m)$,

which simplifies to $\left(\frac{n}{m}\right)^{2}+\frac{n}{m}-1=0$.
Solving this, we get $\frac{n}{m}=\frac{\sqrt{5}-1}{2}$ or $\frac{n}{m}=\frac{-\sqrt{5}-1}{2}$ (discard the latter).
Since Rt $\triangle A F E \backsim \mathrm{Rt} \triangle C F B$, then
$$
\frac{A E}{A D}=\frac{A E}{B C}=\frac{A F}{F C}=\frac{n}{m}=\frac{\sqrt{5}-1}{2} .
$$