连接并延长$CH$交$AD$于点$K$，连接$EK$，作$EL\bot DA$交$DA$的延长线于点$L$，则$\angle L=90^{\circ}$，
$\because$四边形$ABCD$是平行四边形，
$\therefore AD=BC=10,AD$∥$BC$，
$\therefore \angle HDK=\angle HFC$，$\angle HKD=\angle HCF$，
$\because$点$G$、$H$分别是$EC$、$FD$的中点，
$\therefore EG=CG$，$DH=FH$，
在$\triangle HDK$和$\triangle HFC$中，
$\left\{\begin{array}{l}{∠HDK=∠HFC}\\{∠HKD=∠HCF}\\{DH=FH}\end{array}\right.$，
$\therefore \triangle HDK$≌$\triangle HFC\left(AAS\right)$，
$\therefore KH=CH$，$KD=CF$，
$\therefore GH=\frac{1}{2}EK$，
$\because AB=6\sqrt{2}$，点$E$、$F$分别是边$AB$、$BC$的中点，
$\therefore AE=BE=\frac{1}{2}AB=3\sqrt{2}$，$KD=CF=BF=\frac{1}{2}BC=5$，
$\because \angle BAD=135^{\circ}$，
$\therefore \angle LAE=180^{\circ}-\angle BAD=45^{\circ}$，
$\therefore \angle LEA=\angle LAE=45^{\circ}$，
$\therefore EL=AL$，
$\because AE=\sqrt{E{L}^{2}+A{L}^{2}}=\sqrt{2}AL=3\sqrt{2}$，
$\therefore EL=AL=3$，
$\therefore LK=AD+AL-KD=3+10-5=8$，
$\therefore EK=\sqrt{E{L}^{2}+L{K}^{2}}=\sqrt{{3}^{2}+{8}^{2}}=\sqrt{73}$，
$\therefore GH=\frac{1}{2}\times \sqrt{73}=\frac{\sqrt{73}}{2}$，
故答案为：$\frac{\sqrt{73}}{2}$.

连接并延长$CH$交$AD$于点$K$，连接$EK$，作$EL\bot DA$交$DA$的延长线于点$L$，则$\angle L=90^{\circ}$，
$\because$四边形$ABCD$是平行四边形，
$\therefore AD=BC=10,AD$∥$BC$，
$\therefore \angle HDK=\angle HFC$，$\angle HKD=\angle HCF$，
$\because$点$G$、$H$分别是$EC$、$FD$的中点，
$\therefore EG=CG$，$DH=FH$，
在$\triangle HDK$和$\triangle HFC$中，
$\left\{\begin{array}{l}{∠HDK=∠HFC}\\{∠HKD=∠HCF}\\{DH=FH}\end{array}\right.$，
$\therefore \triangle HDK$≌$\triangle HFC\left(AAS\right)$，
$\therefore KH=CH$，$KD=CF$，
$\therefore GH=\frac{1}{2}EK$，
$\because AB=6\sqrt{2}$，点$E$、$F$分别是边$AB$、$BC$的中点，
$\therefore AE=BE=\frac{1}{2}AB=3\sqrt{2}$，$KD=CF=BF=\frac{1}{2}BC=5$，
$\because \angle BAD=135^{\circ}$，
$\therefore \angle LAE=180^{\circ}-\angle BAD=45^{\circ}$，
$\therefore \angle LEA=\angle LAE=45^{\circ}$，
$\therefore EL=AL$，
$\because AE=\sqrt{E{L}^{2}+A{L}^{2}}=\sqrt{2}AL=3\sqrt{2}$，
$\therefore EL=AL=3$，
$\therefore LK=AD+AL-KD=3+10-5=8$，
$\therefore EK=\sqrt{E{L}^{2}+L{K}^{2}}=\sqrt{{3}^{2}+{8}^{2}}=\sqrt{73}$，
$\therefore GH=\frac{1}{2}\times \sqrt{73}=\frac{\sqrt{73}}{2}$，
故答案为：$\frac{\sqrt{73}}{2}$.