证明：连接AE,设AC与DE交于O∵⊿ABC是等边三角形∴∠ACB=60º,则∠ACF=120º∵CE平分∠ACF∴∠ACE=60º=∠ADE又∵∠AOD=∠EOC∴⊿AOD∽⊿EOC（AA‘）∴AO/EO=DO/CO即AO/DO=EO/CO又∵∠AOE=∠DOC∴⊿AOE...