已知等边三角形ABC过C点做直线a平行于AB,D点在CB的延长线上,E点为直线a上的点,且满足角ADE等于60°.证明A

在直线a上取一点F,使得：CF = CA ,且点F和点A在BC的两侧.
已知,AB‖CE,可得：∠FCD = ∠ABC = 60° .
在△ACD和△FCD中,CA = CF ,∠ACD = 60° = ∠FCD ,CD为公共边,
所以,△ACD ≌ △FCD ,可得：AD = DF ,∠CAD = ∠CFD .
∠CDE = ∠ADE-∠ADB = 60°-∠ADB = ∠ABC-∠ADB = ∠BAD .
① 若点F和点E重合,
则有：AD = DE .
② 若点F在CE的延长线上,
则有：∠DEF = ∠FCD+∠CDE = ∠CAB+∠BAD = ∠CAD = ∠CFD ,
可得：DE = DF = AD .
③ 若点F在CE的上,
则有：∠DEF = 360°-(∠ADE+∠ACE)-∠CAD = 180°-∠CAD = 180°-∠CFD = ∠DFE ,
可得：DE = DF = AD .
综上可得：AD = DE .

AD=DE