AC=BC,AC垂直BC,AE垂直CF,BF垂直CF,C,E,F分别为垂足,且角BCF=角ABF,CF交AB于D.判断三

三角形ADC是等腰三角形
AE⊥CF,BF⊥CF → AE‖BF → ∠EAD=∠FBD
AE⊥CF → ∠EAC+∠ACE=90
AC⊥BC → ∠ACE+∠BCD=90
→ ∠EAC=∠BCD=∠BCF=∠EAD
→ △ADE≌△ACE → AD=AC