解：在AC上截取AH=AF，连接OH∵∠A=60°∴∠BAC+∠BCA=120°∵AO、BO是角平分线∴∠OAC+∠OCA=60°∴∠AOC=120°∴∠AOF=60°∵∠OAF=∠OAH，AF=AH，AO=AO∴△AOF≌△AOH∴∠AOH=∠OAF=60°，OF=OH∴∠COH=∠COE=60°∵∠OCH=∠OCE，CO=CO∴△COE≌△COH∴OE=OH∴OE=OF
