(1)證明:因?yàn)?DE\perp AC,$$CE=1,$$DE=2,$所以$CD^{2}=CE^{2}+DE^{2}=1^{2}+2^{2}=5。$因?yàn)?AE=4,$所以$AC=AE + CE=4 + 1=5,$$AD^{2}=AE^{2}+DE^{2}=4^{2}+2^{2}=20。$因?yàn)?AD^{2}+CD^{2}=20 + 5=25=AC^{2},$所以$\angle ADC=90^{\circ}。$
(2)解:因?yàn)?AD$是$\triangle ABC$的中線,所以$BD=DC,$由
(1)知$CD=\sqrt{5},$所以$BD=\sqrt{5}。$在$Rt\triangle ABD$中,$AB^{2}=AD^{2}+BD^{2}=20 + 5=25,$則$AB=5。$因?yàn)?DF$是$\triangle ABD$的中線,所以$DF=\frac{1}{2}AB=\frac{5}{2}。$