证明:连接AO∵BO平分∠ABC,CO平分∠ACB∴O是三角形ABC的内心∴AO平分∠BAC∴∠BAO=∠CAO∵OE∥AC∴∠AOE=∠CAO∴∠AOE=∠BAO∴AE=OE∵OF∥AB∴∠AOF=∠BAO∴∠AOF=∠CAO∴态禅如AF=EF又∵AO=袭咐AO∴△AOE≌△AOF∴AE=AF=OE=OF∴菱形帆启AEOF