Question

EF是BC的垂直平分线,AF,BE交于D,AB=AF,求证AD=DF

Answer 1

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∵AH//EF,∴四边形AHFE为平行四边形,∴AD=DF