证明：
作DG平行AC交BC于G,则DG=1/2 AC,GC=1/2 BC.
由△CEF相似于△GDF,
则：CE/DG=CF/GF
则：CE/(1/2AC)=CF/(GC+CF)
CE/(1/2AE+1/2CE)=CF/(1/2BC+1/2CF+1/2CF)
将1/2销去,
得：CE/(AE+CE)=CF/(BC+CF+CF)
即CE/(AE+CE)=CF/(BF+CF)
即(AE+CE)/CE=(BF+CF)/CF
AE/CE+1=BF/CF+1
所以：BF/CF=AE/EC