过E作EF∥AC交BC于F.
∵EF∥AC,∴△ABC∽△EBF,而△ABC是等边三角形,∴△EBF也是等边三角形,
∴BE=BF=EF、∠EBF=∠EFB=60°,∴∠EBD=∠EFC=120°.
∵△ABC是等边三角形,∴AB=BC,又BE=BF,∴AE=CF.
∵DE=CE,∴∠BDE=∠FCE,结合证得的∠EBD=∠EFC、BE=FE,得:△DBE≌△CFE,
∴DB=CF,结合证得的AE=CF,得:AE=DB.
过E作EF∥AC交BC于F.
∵EF∥AC,∴△ABC∽△EBF,而△ABC是等边三角形,∴△EBF也是等边三角形,
∴BE=BF=EF、∠EBF=∠EFB=60°,∴∠EBD=∠EFC=120°.
∵△ABC是等边三角形,∴AB=BC,又BE=BF,∴AE=CF.
∵DE=CE,∴∠BDE=∠FCE,结合证得的∠EBD=∠EFC、BE=FE,得:△DBE≌△CFE,
∴DB=CF,结合证得的AE=CF,得:AE=DB.