Question

点C为线段AB上一点,△ACM,△CBN是等边三角形,求证:CF平分∠AFB

Answer 1

已知△ACM和△CBN是等边三角形∴AC＝MC,BC＝NC,∠ACN＝∠MCB＝120°,∠AMC＝60°∴△ACN≌MCB (SAS)∴∠NAC＝∠BMC,即∠FAC＝∠FMC∴ACFM四点共圆∴∠AFC＝∠AMC＝60°同理,可得：∠BFC＝∠BNC＝60°∴∠AFC＝∠BFC...