证明:∵等边三角形
∴AB=BC ∠ABC=∠ACB=60°
∵AD=CD∵
∴∠DBE=1/2∠ABC=30°
∵BD=DE
∴∠DBE=∠E=30°
∵∠ACB=∠CDE+∠E
∴∠CDE=∠E=30°
∴CE=CD