已知,如图,正方形ABCD中,点E为BC上一点,AF平分角DAE交CD于F,求证AE=BE+DF（一定要延长CB的那种）

证明：在CB的延长线上取点G,使BG＝DF,连接AG∵正方形ABCD∴AB＝AD,∠ABG＝∠ADC＝90∵BG＝DF∴△ABG≌△ADF （SAS）∴∠G＝∠AFD,∠BAG＝∠DAF∵AF平分∠DAE∴∠DAF＝∠EAF∴∠GAE＝∠BAG+∠BAE＝∠DAF+∠BAE∵AB∥CD∴∠AFD＝∠BAF＝∠EAF+∠BAE＝∠DAF+∠BAE∴∠G＝∠DAF+∠BAE∴∠G＝∠GAE∴AE＝GE∵GE＝BE+BG∴GE＝BE+DF∴AE＝BE+DF