$\because \triangle ABC$和$\triangle DCE$均是等边三角形，
$\therefore BC=AC$，$CD=CE$，$\angle ACB=\angle ECD=60^{\circ}$，
$\therefore \angle ACB+\angle ACD=\angle ACD+\angle ECD$，$\angle ACD=60^{\circ}$，
$\therefore \triangle BCD$≌$\triangle ACE\left(SAS\right)$，
$\therefore AE=BD$，
$\angle CBD=\angle CAE$，
$\because \angle BCA=\angle ACG=60^{\circ}$，$AC=BC$，
$\therefore \triangle BCF$≌$\triangle ACG\left(ASA\right)$
$\therefore AG=BF$，$(①正确)$
同理：$\triangle DFC$≌$\triangle EGC\left(ASA\right)$，
$\therefore CF=CG$，
$\therefore \triangle CFG$是等边三角形，
$\therefore \angle CFG=\angle FCB=60^{\circ}$，
$\therefore FG$∥$BE,(②正确)$
在$DFC$和$\triangle GCE$中
$\left\{\begin{array}{l}{∠FDC=∠GEC}\\{∠DCF=∠ECG}\\{CF=CG}\end{array}\right.$，
$\therefore \triangle DFC$≌$\triangle GCE\left(AAS\right)$，
$\therefore DF=EG$，$DC=CE$，$(③不正确)$
$\because \angle DOE=\angle OBE+\angle OEB$，$\angle OBE=\angle OAC$，
$\therefore \angle DOE=\angle OAC+\angle OEB$，
$\therefore \angle DOE=\angle ACB$，
$\because \angle ACB=60^{\circ}$，
$\therefore \angle DOE=60^{\circ}$，（④正确）
过$C$作$CM\bot AE$于$M$，$CN\bot BD$于$N$，

$\because \triangle BCD$≌$\triangle ACE\left(ASA\right)$，
$\therefore \angle BDC=\angle AEC$，
$\because CD=CE$，$\angle CND=\angle CMA=90^{\circ}$，
$\therefore \triangle CDN$≌$\triangle CEM$，
$\therefore CM=CN$，
$\because CM\bot AE$，$CN\bot BD$，
$\therefore \triangle Rt\triangle OCN$≌$Rt\triangle OCM\left(HL\right)$，
$\therefore \angle BOC=\angle EOC$，
$\therefore$⑤正确；
故答案为：①②④⑤.

$\because \triangle ABC$和$\triangle DCE$均是等边三角形，
$\therefore BC=AC$，$CD=CE$，$\angle ACB=\angle ECD=60^{\circ}$，
$\therefore \angle ACB+\angle ACD=\angle ACD+\angle ECD$，$\angle ACD=60^{\circ}$，
$\therefore \triangle BCD$≌$\triangle ACE\left(SAS\right)$，
$\therefore AE=BD$，
$\angle CBD=\angle CAE$，
$\because \angle BCA=\angle ACG=60^{\circ}$，$AC=BC$，
$\therefore \triangle BCF$≌$\triangle ACG\left(ASA\right)$
$\therefore AG=BF$，$(①正确)$
同理：$\triangle DFC$≌$\triangle EGC\left(ASA\right)$，
$\therefore CF=CG$，
$\therefore \triangle CFG$是等边三角形，
$\therefore \angle CFG=\angle FCB=60^{\circ}$，
$\therefore FG$∥$BE,(②正确)$
在$DFC$和$\triangle GCE$中
$\left\{\begin{array}{l}{∠FDC=∠GEC}\\{∠DCF=∠ECG}\\{CF=CG}\end{array}\right.$，
$\therefore \triangle DFC$≌$\triangle GCE\left(AAS\right)$，
$\therefore DF=EG$，$DC=CE$，$(③不正确)$
$\because \angle DOE=\angle OBE+\angle OEB$，$\angle OBE=\angle OAC$，
$\therefore \angle DOE=\angle OAC+\angle OEB$，
$\therefore \angle DOE=\angle ACB$，
$\because \angle ACB=60^{\circ}$，
$\therefore \angle DOE=60^{\circ}$，（④正确）
过$C$作$CM\bot AE$于$M$，$CN\bot BD$于$N$，

$\because \triangle BCD$≌$\triangle ACE\left(ASA\right)$，
$\therefore \angle BDC=\angle AEC$，
$\because CD=CE$，$\angle CND=\angle CMA=90^{\circ}$，
$\therefore \triangle CDN$≌$\triangle CEM$，
$\therefore CM=CN$，
$\because CM\bot AE$，$CN\bot BD$，
$\therefore \triangle Rt\triangle OCN$≌$Rt\triangle OCM\left(HL\right)$，
$\therefore \angle BOC=\angle EOC$，
$\therefore$⑤正确；
故答案为：①②④⑤.