9. (16 points) As shown in Figure 1, in rectangle $A B C D$, it is known that $A B=2, A D=4$. Points $E$ and $F$ are on $A D$ and $B C$ respectively, and $A E=1, B F=3$. Quadrilateral $A E F B$ is folded along $E F$, such that the projection of point $B$ on plane $C D E F$ is point $H$ on line $D E$. Find the size of the dihedral angle $A-D E-F$.
9. As shown in Figure 3, draw $E R / / D C$, and draw $E S \perp$ plane $E F C D$. Establish a spatial rectangular coordinate system with $E R$, $E D$, and $E S$ as the $x$, $y$, and $z$ axes, respectively.

Notice that the projection of point $B$ on plane $C D E F$ is point $H$ on line $D E$. Let point $B(0, y, z)(y>0, z>0)$.
Since $F(2,2,0)$, $B E=\sqrt{5}$, and $B F=3$, we have:
$$
\left\{\begin{array}{l}
y^{2}+z^{2}=5, \\
4+(y-2)^{2}+z^{2}=9
\end{array} \Rightarrow\left\{\begin{array}{l}
y=1, \\
z=2 .
\end{array}\right.\right.
$$
Thus, $B(0,1,2)$.
Therefore, $\overrightarrow{F B}=(-2,-1,2)$.
Hence, $\overrightarrow{E A}=\frac{1}{3} \overrightarrow{F B}=\left(-\frac{2}{3},-\frac{1}{3}, \frac{2}{3}\right)$.
Let a normal vector of plane $A D E$ be $\boldsymbol{n}=(x, y, z)$.
Also, $\overrightarrow{E D}=(0,3,0)$. Then
$$
\left\{\begin{array}{l}
-\frac{2}{3} x-\frac{1}{3} y+\frac{2}{3} z=0, \\
3 y=0
\end{array} \Rightarrow n=(1,0,1) .\right.
$$

Notice that a normal vector of plane $C D E F$ is $m=(0,0,1)$.
Let the dihedral angle $A-D E-F$ be $\theta$.
Clearly, $\theta$ is an obtuse angle.
Then $\cos \theta=-|\cos |=-\frac{\sqrt{2}}{2}$.
Thus, $\theta=135^{\circ}$.