①$\because \angle A=30^{\circ}$，
$\therefore BC=\frac{1}{2}AC$，
$\because$点$D$是$AC$的中点，
$\therefore BD=\frac{1}{2}AC$，
$\therefore AD=CD=BC=BD$，
$\therefore \triangle BCD$是等腰三角形，
$\because$点$F$是$CD$的中点，
$\therefore BF\bot CD$，
$\therefore \angle BFC=90^{\circ}$，
$\because \angle A+\angle C=90^{\circ}$，$\angle FBC+\angle C=90^{\circ}$，
$\therefore \angle A=\angle FBC=30^{\circ}$，
$\therefore \angle FBE=\angle ABC-\angle FBC=60^{\circ}$，
$\because D$，$E$分别是$AC$、$AB$的中点，
$\therefore DE=\frac{1}{2}BC=\frac{1}{4}AC,DE$∥$BC$，
$\because \angle ABC=90^{\circ}$，
$\therefore \angle BED=90^{\circ}$，
$\because$点$F$是$CD$的中点，
$\therefore DF=\frac{1}{2}CD=\frac{1}{4}AC$，
在$Rt\triangle BED$和$Rt\triangle BFD$中，
$\left\{\begin{array}{l}{DE=DF}\\{BD=BD}\end{array}\right.$，
$\therefore Rt\triangle BED$≌$Rt\triangle BFD\left(HL\right)$，
$\therefore BE=BF$，
$\therefore \triangle EFB$是等腰三角形，
$\because \angle FBE=60^{\circ}$，
$\therefore \triangle EFB$是等边三角形，
$\therefore$①正确.
②$\because Rt\triangle BED$≌$Rt\triangle BFD$，
$\therefore S_{\triangle BED}=S_{\triangle BFD}$，
$\because AE=BE$，
$\therefore S_{\triangle AED}=S_{\triangle BED}$，
$\because$点$F$是$CD$的中点，
$\therefore S_{\triangle BFD}=S_{\triangle BFC}$，
$\therefore S_{\triangle AED}=S_{\triangle BED}=S_{\triangle BFD}=S_{\triangle BFC}$，
$\therefore S_{四边形DFBE}=S_{\triangle BED}+S_{\triangle BFD}=\frac{1}{2}S_{\triangle ABC}$，
$\therefore$②正确.
③$\because AB=AC\cdot \cos \angle A=\frac{\sqrt{3}}{2}AC$，
$\therefore AE=\frac{1}{2}AB=\frac{\sqrt{3}}{4}AC$，
$\because DF=\frac{1}{4}AC$，
$\therefore AE=\sqrt{3}DF$，
$\therefore$③不正确.
④$\because DE=DF$，$BE=BF$，
$\therefore BD$是$EF$的垂直平分线，
$\therefore \angle DGE=90^{\circ}$，
$\because \angle DGE=\angle DEB=90^{\circ}$，$\angle EDG=\angle BDE$，
$\therefore Rt\triangle DGE$∽$Rt\triangle DEB$，
$\therefore \frac{DG}{DE}=\frac{DE}{BD}$，
$\therefore DG=\frac{D{E}^{2}}{BD}=\frac{(\frac{1}{4}AC)^{2}}{\frac{1}{2}AC}=\frac{AC}{8}$，
$\therefore AC=8DG$，
$\therefore$④正确.
故答案为：①②④.

①$\because \angle A=30^{\circ}$，
$\therefore BC=\frac{1}{2}AC$，
$\because$点$D$是$AC$的中点，
$\therefore BD=\frac{1}{2}AC$，
$\therefore AD=CD=BC=BD$，
$\therefore \triangle BCD$是等腰三角形，
$\because$点$F$是$CD$的中点，
$\therefore BF\bot CD$，
$\therefore \angle BFC=90^{\circ}$，
$\because \angle A+\angle C=90^{\circ}$，$\angle FBC+\angle C=90^{\circ}$，
$\therefore \angle A=\angle FBC=30^{\circ}$，
$\therefore \angle FBE=\angle ABC-\angle FBC=60^{\circ}$，
$\because D$，$E$分别是$AC$、$AB$的中点，
$\therefore DE=\frac{1}{2}BC=\frac{1}{4}AC,DE$∥$BC$，
$\because \angle ABC=90^{\circ}$，
$\therefore \angle BED=90^{\circ}$，
$\because$点$F$是$CD$的中点，
$\therefore DF=\frac{1}{2}CD=\frac{1}{4}AC$，
在$Rt\triangle BED$和$Rt\triangle BFD$中，
$\left\{\begin{array}{l}{DE=DF}\\{BD=BD}\end{array}\right.$，
$\therefore Rt\triangle BED$≌$Rt\triangle BFD\left(HL\right)$，
$\therefore BE=BF$，
$\therefore \triangle EFB$是等腰三角形，
$\because \angle FBE=60^{\circ}$，
$\therefore \triangle EFB$是等边三角形，
$\therefore$①正确.
②$\because Rt\triangle BED$≌$Rt\triangle BFD$，
$\therefore S_{\triangle BED}=S_{\triangle BFD}$，
$\because AE=BE$，
$\therefore S_{\triangle AED}=S_{\triangle BED}$，
$\because$点$F$是$CD$的中点，
$\therefore S_{\triangle BFD}=S_{\triangle BFC}$，
$\therefore S_{\triangle AED}=S_{\triangle BED}=S_{\triangle BFD}=S_{\triangle BFC}$，
$\therefore S_{四边形DFBE}=S_{\triangle BED}+S_{\triangle BFD}=\frac{1}{2}S_{\triangle ABC}$，
$\therefore$②正确.
③$\because AB=AC\cdot \cos \angle A=\frac{\sqrt{3}}{2}AC$，
$\therefore AE=\frac{1}{2}AB=\frac{\sqrt{3}}{4}AC$，
$\because DF=\frac{1}{4}AC$，
$\therefore AE=\sqrt{3}DF$，
$\therefore$③不正确.
④$\because DE=DF$，$BE=BF$，
$\therefore BD$是$EF$的垂直平分线，
$\therefore \angle DGE=90^{\circ}$，
$\because \angle DGE=\angle DEB=90^{\circ}$，$\angle EDG=\angle BDE$，
$\therefore Rt\triangle DGE$∽$Rt\triangle DEB$，
$\therefore \frac{DG}{DE}=\frac{DE}{BD}$，
$\therefore DG=\frac{D{E}^{2}}{BD}=\frac{(\frac{1}{4}AC)^{2}}{\frac{1}{2}AC}=\frac{AC}{8}$，
$\therefore AC=8DG$，
$\therefore$④正确.
故答案为：①②④.