延长EF交AD于G点∵OB=OD,∠CBD=∠ADB,∠BOF=∠DOF∴△BOF≌△DOG∴DG=BF,可知:AG=AD-DG=BC-BF=c-BF∵BC//AD∴△EFB∽△EGA∴BF/AG=BE/AE,即:BF/(c-BF)=b/(a+b),可解得:BF=bc/(a+2b)
