EF是BC的垂直平分线,AF,BE交于D,AB=AF,求证AD=DF

过A作AG//EF,分别交BE,BC于H,G两点.再连结HF∵EF⊥BC,AG//EC,∴AG⊥BC,而AB=AF,故AG是三角形ABF的中垂线.∴BH=FH,∴∠HBC=∠HFB再由EF是BC的垂直平分线得到,BE=CE,∴∠ECB=∠HBC,∴∠ECB=∠HFB,∴HF//AC,即AE//HF.又...