(1)证明:在AC上截取AF=AE,
因为:AD平分∠BAC,
故:∠EAO=∠FAO=1/2∠BAC,AO=AO
故:△EAO≌△FAO
故:∠EOA=∠FOA
CE平分∠ACB
故:∠ACO=∠BCO=1/2∠ACB
又:∠BAC+∠ACB+∠B=180°,∠B=60°
所以:∠BAC+∠ACB=120°
所以:∠EOA=∠FOA=∠FAO+∠ACO=1/2(∠BAC+∠ACB)=60°
所以:∠COF=60°=∠EOA=∠DOC
又:∠ACO=∠BCO ,OC=OC
故:△COF≌△COD
故:CD=CF
故:AC=AF+CF=AE+CD
(2)由:△EAO≌△FAO
得:OE=OF
再由:△COF≌△COD
得:OD=OF
所以OE=OD