6. 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 $GH=$
6. $\frac{4}{9} \sqrt{21}$ Detailed explanation: Let the plane $A H D$ intersect $B C$ at $F$, then because $A B=A C$, $G$ lies on $A F$ and $G F=\frac{1}{3} A F$, $\angle A F H=60^{\circ}$. Therefore, $A F=\frac{A H}{\sin 60^{\circ}}=\frac{8}{\sqrt{3}}, F H=\frac{1}{2} A F=\frac{4}{\sqrt{3}}, G F=\frac{8}{3 \sqrt{3}}$. In triangle $G F H$, by the cosine rule, $G H=\frac{4}{9} \sqrt{21}$.