Question

如图,正方形ABCD的边BC上有一点E,∠DAE得平分线交CD与F 求证 AE＝DF＋BE

Answer 1

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∥C...