It is proved that: (1) ∠ ACB = 90, the midpoint of E line segment AB.

∴CE= 12AB=AE，

∫∠ACD = 90, f is the midpoint of line segment AD,

∴AF=CF= 12AD，

At △CEF and △AEF,

CF=AFEF=EFCE=AE，

∴△cef≌△aef(sss)；

(2) the connection DE,

Points e and f are the midpoints of line segments AB and AD, respectively,

∴EF= 12BD,EF∥BC，

BD = 2CD，

∴EF=CD.

And ∵EF∨BC,

∴ quadrilateral CEFD is a parallelogram,

∴DE=CF，

CF = AF，

∴DE=AF.