方法一：连接$CH$并延长交$AD$于$P$，连接$PE$，

$\because$四边形$ABCD$是正方形，
$\therefore \angle A=90^{\circ}$，$AD$∥$BC$，$AB=AD=BC=2\sqrt{2}$，
$\because E$，$F$分别是边$AB$，$BC$的中点，
$\therefore AE=CF=\frac{1}{2}\times 2\sqrt{2}=\sqrt{2}$，
$\because AD$∥$BC$，
$\therefore \angle DPH=\angle FCH$，
$\because \angle DHP=\angle FHC$，
$\because DH=FH$，
$\therefore \triangle PDH$≌$\triangle CFH\left(AAS\right)$，
$\therefore PD=CF=\sqrt{2}$，
$\therefore AP=AD-PD=\sqrt{2}$，
$\therefore PE=\sqrt{A{P}^{2}+A{E}^{2}}=\sqrt{(\sqrt{2})^{2}+(\sqrt{2})^{2}}=2$，
$\because$点$G$，$H$分别是$EC$，$CP$的中点，
$\therefore GH=\frac{1}{2}EP=1$；
方法二：设$DF$，$CE$交于$O$，
$\because$四边形$ABCD$是正方形，
$\therefore \angle B=\angle DCF=90^{\circ}$，$BC=CD=AB$，
$\because$点$E$，$F$分别是边$AB$，$BC$的中点，
$\therefore BE=CF$，
$\therefore \triangle CBE$≌$\triangle DCF\left(SAS\right)$，
$\therefore CE=DF$，$\angle BCE=\angle CDF$，
$\because \angle CDF+\angle CFD=90^{\circ}$，
$\therefore \angle BCE+\angle CFD=90^{\circ}$，
$\therefore \angle COF=90^{\circ}$，
$\therefore DF\bot CE$，
$\therefore CE=DF=\sqrt{(2\sqrt{2})^{2}+(\sqrt{2})^{2}}=\sqrt{10}$，
$\because$点$G$，$H$分别是$EC$，$PC$的中点，
$\therefore CG=FH=\frac{\sqrt{10}}{2}$，
$\because \angle DCF=90^{\circ}$，$CO\bot DF$，
$\therefore \angle DCO+\angle FCO=\angle DCO+\angle CDO=90^{\circ}$，
$\therefore \angle FCO=\angle CDO$，
$\because \angle DCF=\angle COF=90^{\circ}$，
$\therefore \triangle COF$∽$\triangle DOC$，
$\therefore \frac{CF}{DF}=\frac{OF}{CF}$，
$\therefore CF^{2}=OF\cdot DF$，
$\therefore OF=\frac{C{F}^{2}}{DF}=\frac{(\sqrt{2})^{2}}{\sqrt{10}}=\frac{\sqrt{10}}{5}$，
$\therefore OH=\frac{3\sqrt{10}}{10}$，$OD=\frac{4\sqrt{10}}{5}$，
$\because \angle COF=\angle COD=90^{\circ}$，
$\therefore \triangle COF$∽$\triangle DCF$，
$\therefore \frac{OF}{OC}=\frac{OC}{OD}$，
$\therefore OC^{2}=OF\cdot OD$，
$\therefore OC=\sqrt{\frac{\sqrt{10}}{5}×\frac{4\sqrt{10}}{5}}=\frac{2\sqrt{10}}{5}$，
$\therefore OG=CG-OC=\frac{\sqrt{10}}{2}-\frac{2\sqrt{10}}{5}=\frac{\sqrt{10}}{10}$，
$\therefore HG=\sqrt{O{G}^{2}+O{H}^{2}}=\sqrt{\frac{1}{10}+\frac{9}{10}}=1$，
故答案为：$1$.

方法一：连接$CH$并延长交$AD$于$P$，连接$PE$，

$\because$四边形$ABCD$是正方形，
$\therefore \angle A=90^{\circ}$，$AD$∥$BC$，$AB=AD=BC=2\sqrt{2}$，
$\because E$，$F$分别是边$AB$，$BC$的中点，
$\therefore AE=CF=\frac{1}{2}\times 2\sqrt{2}=\sqrt{2}$，
$\because AD$∥$BC$，
$\therefore \angle DPH=\angle FCH$，
$\because \angle DHP=\angle FHC$，
$\because DH=FH$，
$\therefore \triangle PDH$≌$\triangle CFH\left(AAS\right)$，
$\therefore PD=CF=\sqrt{2}$，
$\therefore AP=AD-PD=\sqrt{2}$，
$\therefore PE=\sqrt{A{P}^{2}+A{E}^{2}}=\sqrt{(\sqrt{2})^{2}+(\sqrt{2})^{2}}=2$，
$\because$点$G$，$H$分别是$EC$，$CP$的中点，
$\therefore GH=\frac{1}{2}EP=1$；
方法二：设$DF$，$CE$交于$O$，
$\because$四边形$ABCD$是正方形，
$\therefore \angle B=\angle DCF=90^{\circ}$，$BC=CD=AB$，
$\because$点$E$，$F$分别是边$AB$，$BC$的中点，
$\therefore BE=CF$，
$\therefore \triangle CBE$≌$\triangle DCF\left(SAS\right)$，
$\therefore CE=DF$，$\angle BCE=\angle CDF$，
$\because \angle CDF+\angle CFD=90^{\circ}$，
$\therefore \angle BCE+\angle CFD=90^{\circ}$，
$\therefore \angle COF=90^{\circ}$，
$\therefore DF\bot CE$，
$\therefore CE=DF=\sqrt{(2\sqrt{2})^{2}+(\sqrt{2})^{2}}=\sqrt{10}$，
$\because$点$G$，$H$分别是$EC$，$PC$的中点，
$\therefore CG=FH=\frac{\sqrt{10}}{2}$，
$\because \angle DCF=90^{\circ}$，$CO\bot DF$，
$\therefore \angle DCO+\angle FCO=\angle DCO+\angle CDO=90^{\circ}$，
$\therefore \angle FCO=\angle CDO$，
$\because \angle DCF=\angle COF=90^{\circ}$，
$\therefore \triangle COF$∽$\triangle DOC$，
$\therefore \frac{CF}{DF}=\frac{OF}{CF}$，
$\therefore CF^{2}=OF\cdot DF$，
$\therefore OF=\frac{C{F}^{2}}{DF}=\frac{(\sqrt{2})^{2}}{\sqrt{10}}=\frac{\sqrt{10}}{5}$，
$\therefore OH=\frac{3\sqrt{10}}{10}$，$OD=\frac{4\sqrt{10}}{5}$，
$\because \angle COF=\angle COD=90^{\circ}$，
$\therefore \triangle COF$∽$\triangle DCF$，
$\therefore \frac{OF}{OC}=\frac{OC}{OD}$，
$\therefore OC^{2}=OF\cdot OD$，
$\therefore OC=\sqrt{\frac{\sqrt{10}}{5}×\frac{4\sqrt{10}}{5}}=\frac{2\sqrt{10}}{5}$，
$\therefore OG=CG-OC=\frac{\sqrt{10}}{2}-\frac{2\sqrt{10}}{5}=\frac{\sqrt{10}}{10}$，
$\therefore HG=\sqrt{O{G}^{2}+O{H}^{2}}=\sqrt{\frac{1}{10}+\frac{9}{10}}=1$，
故答案为：$1$.