(1) 延长EA至点G,使得AG=BE,连接DG
则由AG=BE,AD=AE,∠DAG=90°=∠AEB
∴△ADG≌△EAB,∴AB=DG
∠GFD=90°-∠ADF
=90°-∠ADC/2
=90°-∠ABC/2
=90°-∠ABC+∠ABC/2
=∠BAE+∠ADC/2
=∠GDA+∠ADF
=∠GDF
∴GF=GD
而CD=AB=GD,GF=GA+AF=BE+AF
∴CD=AF+BE
(2)若BE=AF,则AG=AF
∴DA即是垂线也是中线
∴DG=DF,即△DGF是等边△
∴EF=AE-AF=AD-AF=(√3-1)AF