11. As shown in the figure, in the quadrilateral pyramid $S-ABCD$, $SD \perp$ the base $ABCD$, $AB \| DC$, $AD \perp DC$, $AB=AD=1$, $DC=SD=2$, $E$ is a point on the edge $SB$, and the plane $EDC \perp$ the plane $SBC$.
(1) Prove: $SE=2EB$;
(2) Find the size of the dihedral angle $A-DE-C$.
Parse: With $D$ as the origin, the rays $D A, D C, D S$ are the $x$-axis, $y$-axis, $z$-axis respectively, establishing the rectangular coordinate system $D x y z$. Let $A=(1,0,0)$, then $B(1,1,0), C(0,2,0), S(0,0,2)$.
5 points
Proof (1) $\overrightarrow{S C}=(0,2,-2), \overrightarrow{B C}=(-1,1,0)$.
Let the normal vector of plane $S B C$ be $n=(a, b, c)$. From $n \perp \overrightarrow{S C}, n \perp \overrightarrow{B C}$, we get $n \cdot \overrightarrow{S C}=0, n \cdot \overrightarrow{B C}=0$, hence $b-c=0,-a+b=0$. Taking $a=b=c=1$, then $n=(1,1,1)$. Also, let $\overrightarrow{S E}=\lambda \overrightarrow{E B}(\lambda>0)$, then
$$
E\left(\frac{\lambda}{1+\lambda}, \frac{\lambda}{1+\lambda}, \frac{2}{1+\lambda}\right), \quad \overrightarrow{D E}=\left(\frac{\lambda}{1+\lambda}, \frac{\lambda}{1+\lambda}, \frac{2}{1+\lambda}\right), \quad \overrightarrow{D C}=(0,2,0)
$$

Let the normal vector of plane $C D E$ be $m=(x, y, z)$. From $m \perp \overrightarrow{D E}, m \perp \overrightarrow{D C}$, we get $m \cdot \overrightarrow{D E}=0, m \cdot \overrightarrow{D C}=0$, hence
$$
\frac{\lambda x}{1+\lambda}+\frac{\lambda y}{1+\lambda}+\frac{2 z}{1+\lambda}=0, \quad 2 y=0
$$

Let $x=2$, then $m=(2,0,-\lambda)$, from plane $D E C \perp$ plane $S B C$, we get $m \perp n$, so
$$
m \cdot n=0,2-\lambda=0, \quad \lambda=2
$$

Thus, $S E=2 E B$.
$$
\overrightarrow{F A} \cdot \overrightarrow{D E}=0 \Rightarrow \overrightarrow{F A} \perp \overrightarrow{D E}
$$

Also, $\overrightarrow{E C}=\left(-\frac{2}{3}, \frac{4}{3},-\frac{2}{3}\right)$, hence $\overrightarrow{E C} \perp \overrightarrow{D E}$, therefore the angle between vector $\overrightarrow{F A}$ and $\overrightarrow{E C}$ equals the plane angle of the dihedral angle $A-D E-C$, thus
$$
\cos (\overrightarrow{F A}, \overrightarrow{E C})=\frac{\overrightarrow{F A} \cdot \overrightarrow{E C}}{|\overrightarrow{F A}||\overrightarrow{E C}|}=-\frac{1}{2}
$$

Therefore, the size of the dihedral angle $A-D E-C$ is $120^{\circ}$.