过点A作AM⊥BE于M,AN⊥CD于N∵∠BAD＝60,AB＝AD∴等边△ABD∴∠ABD＝∠ADB＝60∵∠BAE＝∠BAC+∠CAE,∠DAC＝∠BAC+∠BAD,∠BAD＝∠CAE∴∠BAE＝∠DAC∵AC＝AE∴△BAE≌△DAC （SAS）∴∠ABE＝∠ADC,BE＝CD,S△BAE＝S...BAD=90度时呢，只要答案∠ADB+∠ABD＝90则∠DFE＝90∠AFE＝45