(9) In tetrahedron $ABCD$, face $ABC$ and face $BCD$ form a dihedral angle of $60^{\circ}$, the projection of vertex $A$ onto face $BCD$ is the orthocenter $H$ of $\triangle BCD$, $G$ is the centroid of $\triangle ABC$, if $AH=4$, $AB=AC$, then $GH=$ $\qquad$
9 Let plane $AHD$ intersect $BC$ at point $F$. Since $AB=AC$, point $G$ lies on $AF$, and $GF=\frac{1}{3} AF$. Also, since $\angle AFH=60^{\circ}$, we have
$$
\begin{aligned}
AF & =\frac{AH}{\sin 60^{\circ}}=\frac{8}{\sqrt{3}}, \\
FH & =\frac{1}{2} AF=\frac{4}{\sqrt{3}}, \\
GF & =\frac{8}{3 \sqrt{3}}
\end{aligned}
$$

Therefore, in triangle $GFH$, by the cosine rule, $GH=\frac{4}{9} \sqrt{21}$.