连接AH,DH,延长AD交CF于点G∵△ABC和△DEF均为正三角形,BC和EF的中点均为H∴AH⊥BC,DH⊥EF,AH=√3CH,DH=√3FH∴∠AHD=∠CHF,AH：DH=CH: FH∴△AHD∽△CHF∴∠HAD=∠HCF∵∠HAD+∠DAC+∠ACH=90°∴∠HCF+∠DAC+∠ACH=...