在BC边上取点F,使BF=BE,连结OF.
∵BD是角平分线,∠EBO=∠FBO,BF=BE,BO=BO
∴△BEO≌△BFO,∴∠EOB=∠FOB=∠COD
∵∠A=60°
∠EOB=∠CBO＋∠BCO,又BD、CE是角平分线
∴∠EOB=1/2(180°－60°)=60°
则∠COF=180°－∠FOB－∠COD=60°=∠COD
又CE是角平分线,∴∠OCD=∠OCF,CO=CO,
∴△COF≌△COD ,∴CF＝CD
∴BC＝BF＋CF＝BE＋CD