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 $\triangle BCD$, and $G$ is the centroid of $\triangle ABC$. If $AH=4$ and $AB=AC$, then $GH=$ $\qquad$
9. $\frac{4 \sqrt{21}}{9}$.

Let the plane $A H D$ intersect $B C$ at point $F$. Given $A B=A C$, we know point $G$ lies on $A F$, and
$$
G F=\frac{1}{3} A F, \angle A F H=60^{\circ} \text {. }
$$

Then $A F=\frac{A H}{\sin 60^{\circ}}=\frac{8}{\sqrt{3}}, F H=\frac{1}{2} A F=\frac{4}{\sqrt{3}}$,
$$
G F=\frac{8}{3 \sqrt{3}} .
$$

In $\triangle G F H$, by the cosine rule we get
$$
G H=\frac{4 \sqrt{21}}{9} .
$$