角BAC=角ACG,AB=AC,角EAD与角EDA互余,角EAD与角AGC互余,所以角EDA=AGC,所以△ABD≌△ACG推出AD=DC=CG;CF=CF,∠ACB=∠CGF,∴△DFC≌△FGC 所以∠FDC=∠FGC=∠BAD