18 Given a regular triangular prism $A B C-A_{1} B_{1} C_{1}$, where $A A_{1}=2 A C=4$, extend $C B$ to $D$ such that $C B=B D$.
(1) Prove that line $C_{1} B$ // plane $A B_{1} D$;
(2) Find the sine value of the angle formed by plane $A B_{1} D$ and plane $A C B$.
18 (1) Connect $C_{1} B$, then $C_{1} B_{1}=C B=D B$, and since $C_{1} B_{1} // B D$, quadrilateral $C_{1} B D B_{1}$ is a parallelogram, $C_{1} B // B_{1} D$. Since $B_{1} D \subset$ plane $A B_{1} D$, line $C_{1} B //$ plane $A B_{1} D$.
(2) In $\triangle A C D$, since $C B=B D=B A$, $\angle D A C=90^{\circ}$. Establish a spatial rectangular coordinate system with $A$ as the origin, then $A(0,0,0) 、 B_{1}(\sqrt{3}, 1,4) 、 D(2 \sqrt{3}, 0,0)$, $\overrightarrow{A D}=(2 \sqrt{3}, 0,0), \overrightarrow{A B_{1}}=(\sqrt{3}, 1,4)$. Let the normal vector of plane $A B_{1} D$ be $\vec{n}=(x, y, z)$, then
$$
\left\{\begin{array}{l}
\vec{n} \cdot \overrightarrow{A D}=0 \\
\vec{n} \cdot \overrightarrow{A B_{1}}=0
\end{array}\right.
$$

which is
$$
\left\{\begin{array}{l}
2 \sqrt{3} x=0 \\
\sqrt{3} x+y+4 z=0
\end{array}\right.
$$

So
$$
\left\{\begin{array}{l}
x=0 \\
y=-4 z
\end{array}\right.
$$

Take $z=1$, then $\vec{n}=(0,-4,1)$. Take the normal vector of plane $A C B$ as $\vec{m}=(0,0,1)$, then
$$
\cos \langle\vec{n}, \vec{m}\rangle=\frac{\vec{n} \cdot \vec{m}}{|\vec{n}||\vec{m}|}=\frac{1}{\sqrt{17}}
$$

So $\sin \langle\vec{n}, \vec{m}\rangle=\frac{4 \sqrt{17}}{17}$.
Therefore, the sine value of the angle formed by plane $A B_{1} D$ and plane $A C B$ is $\frac{4 \sqrt{17}}{17}$.