∠AFD和∠AFE相等证明如下：由AD=AB,∠DAB+∠BAC=∠EAC+∠BAC,即∠DAC=∠BAEAC=AE,(边角边)可证明ΔADC≌ΔABE由此推出∠ADF=∠ABE,即∠ADG=∠FBG（假设AB和DF交点是点G）又∠AGD=∠FGB(对顶角相等)则ΔAGD∽ΔFGB（...