9. As shown in the figure, the right trapezoid $A B C D, A B=A D=\frac{1}{2} D C=1$. Fold $\triangle A B D$ along $B D$ so that plane $A B D \perp$ plane $B C D$. Let $G$ be the midpoint of $A D$, $A H=2 H C$, and the midpoint of $B D$ be $O$.
(1) Prove that $A O \perp$ plane $B C D$;
(2) Find the cosine of the acute dihedral angle between plane $G H B$ and plane $B C D$;
(3) Does there exist a point $E$ on segment $B C$ such that $D E / /$ plane $G H B$? If it exists, determine the position of point $E$. If not, explain the reason.
(1) $A B=A D \Rightarrow A O \perp B D$, and $A O \subset$ plane $A B D$, plane $A B D \bigcap$ plane $B C D=B D$, then plane $A B D \perp$ plane $B C D \Rightarrow A O \perp$ plane $B C D$.
(2) $B D=B C=\sqrt{2}, D C=2 \Rightarrow B D \perp B C$, let the midpoint of $D C$ be $F$, establish a rectangular coordinate system as shown in the figure, then $B\left(-\frac{\sqrt{2}}{2}, 0,0\right), D\left(\frac{\sqrt{2}}{2}, 0,0\right), C\left(-\frac{\sqrt{2}}{2}, \sqrt{2}, 0\right), A\left(0,0, \frac{\sqrt{2}}{2}\right) \Rightarrow G\left(\frac{\sqrt{2}}{4}, 0, \frac{\sqrt{2}}{4}\right)$, $H\left(-\frac{\sqrt{2}}{3}, \frac{2 \sqrt{2}}{3}, \frac{\sqrt{2}}{6}\right) \Rightarrow \overrightarrow{B G}=\left(\frac{3 \sqrt{2}}{4}, 0, \frac{\sqrt{2}}{4}\right), \overrightarrow{B H}=\left(\frac{\sqrt{2}}{6}, \frac{2 \sqrt{2}}{3}, \frac{\sqrt{2}}{6}\right)$.
Let the normal vector of plane $B G H$ be $\boldsymbol{n}=(x, y, z)$, then $\left\{\begin{array}{l}3 x+z=0, \\ x+4 y+z=0\end{array} \Rightarrow \boldsymbol{n}=(2,1,-6)\right.$.
Therefore, the cosine value of the acute dihedral angle between plane $G H B$ and plane $B C D$ is $\cos \theta=\frac{|(2,1,-6) \cdot(0,0,1)|}{\sqrt{4+1+36}}=\frac{6 \sqrt{41}}{41}$.
(3) Let a point $E\left(-\frac{\sqrt{2}}{2}, y_{0}, 0\right)\left(0 \leqslant y_{0} \leqslant \sqrt{2}\right)$ on line segment $B C \Rightarrow \overrightarrow{D E}=\left(-\sqrt{2}, y_{0}, 0\right)$, by $\overrightarrow{D E} \cdot \boldsymbol{n}=0 \Rightarrow-2 \sqrt{2}+y_{0}=0 \Rightarrow y_{0}=2 \sqrt{2}>\sqrt{2}$, so there is no point $E$ that satisfies the condition.