14. (13 points) As shown in Figure 1, in the right trapezoid $ABCD$, $AD \parallel BC$, $\angle BAD = \frac{\pi}{2}$, $AB = BC = 1$, $AD = 2$, $E$ is the midpoint of $AD$, and $O$ is the intersection of $AC$ and $BE$. $\triangle ABE$ is folded along $BE$ to the position of $\triangle A_1BE$, as shown in Figure 2.
(1) Prove: $CD \perp$ plane $A_1OC$;
(2) If plane $A_1BE \perp$ plane $BCDE$, find the sine value of the dihedral angle $B-A_1C-D$.
14. (1) In Figure 1, since $A B=B C=1, A D=2, E$ is the midpoint of $A D$, $\angle B A D=\frac{\pi}{2}$, therefore, $B E \perp A C$, which means in Figure 2, $B E \perp O A_{1}, B E \perp O C$, and $A_{1} O$ intersects $O C$ at point $O$.
Thus, $B E \perp$ plane $A_{1} O C$.
Also, since $A D / / B C, B C=\frac{1}{2} A D, E$ is the midpoint of $A D$, then $B C / / E D$.
Therefore, quadrilateral $B C D E$ is a parallelogram.
Hence, $C D / / B E$.
Thus, $C D \perp$ plane $A_{1} O C$.
(2) Given that plane $A_{1} B E \perp$ plane $B C D E$, combined with (1), we know $B E \perp O A_{1}, B E \perp O C$.

Therefore, $\angle A_{1} O C$ is the plane angle of the dihedral angle $A_{1}-B E-C$.
Thus, $\angle A_{1} O C=\frac{\pi}{2}$.
As shown in Figure 6, establish a spatial rectangular coordinate system with $O$ as the origin.

Given $A_{1} B=A_{1} E=B C=E D=1, B C / / E D$, we have
Figure 6
$B\left(\frac{\sqrt{2}}{2}, 0,0\right), E\left(-\frac{\sqrt{2}}{2}, 0,0\right), A_{1}\left(0,0, \frac{\sqrt{2}}{2}\right)$,
$C\left(0, \frac{\sqrt{2}}{2}, 0\right)$.
Then $\overrightarrow{B C}=\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, 0\right)$,
$\overrightarrow{A_{1} C}=\left(0, \frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)$,
$\overrightarrow{C D}=\overrightarrow{B E}=(-\sqrt{2}, 0,0)$.
Let the normal vector of plane $A_{1} B C$ be $\boldsymbol{n}_{1}=\left(x_{1}, y_{1}, z_{1}\right)$, and the normal vector of plane $A_{1} C D$ be $\boldsymbol{n}_{2}=\left(x_{2}, y_{2}, z_{2}\right)$.
Then $\left\{\begin{array}{l}\boldsymbol{n}_{1} \cdot \overrightarrow{B C}=0, \\ \boldsymbol{n}_{1} \cdot \overrightarrow{A_{1} C}=0\end{array} \Rightarrow\left\{\begin{array}{l}-x_{1}+y_{1}=0, \\ y_{1}-z_{1}=0,\end{array}\right.\right.$
Take $\boldsymbol{n}_{1}=(1,1,1)$;
$\left\{\begin{array}{l}\boldsymbol{n}_{2} \cdot \overrightarrow{C D}=0, \\ \boldsymbol{n}_{2} \cdot \overrightarrow{A_{1} C}=0\end{array} \Rightarrow\left\{\begin{array}{l}x_{2}=0, \\ y_{2}-z_{2}=0,\end{array}\right.\right.$
Take $\boldsymbol{n}_{2}=(0,1,1)$.
Thus, $\left|\cos \right|=\frac{2}{\sqrt{3} \times \sqrt{2}}=\frac{\sqrt{6}}{3}$.
Let the dihedral angle $B-A_{1} C-D$ be $\theta$. Then
$$
\sin \theta=\frac{\sqrt{3}}{3} \text {, }
$$

Therefore, the sine value of the dihedral angle $B-A_{1} C-D$ is $\frac{\sqrt{3}}{3}$.