（1）证明：因为$AC\perp BD，$所以$\angle ACB=\angle ACD = 90^{\circ}，$即$\angle CAD+\angle ADC = 90^{\circ}。$
因为$\angle CAD = 45^{\circ}，$所以$\angle ADC = 90^{\circ}-\angle CAD = 45^{\circ}，$即$\angle CAD=\angle ADC。$
所以$AC = DC。$
又$AB = DE，$所以$Rt\triangle ABC\cong Rt\triangle DEC(HL)。$
所以$\angle BAC=\angle EDC。$
因为$\angle EDC+\angle CED = 90^{\circ}，$$\angle CED=\angle AEF，$所以$\angle AEF+\angle BAC = 90^{\circ}。$
又$\angle AEF+\angle BAC+\angle AFE = 180^{\circ}，$所以$\angle AFE = 90^{\circ}。$所以$DF\perp AB。$
（2）证明：由（1），得$Rt\triangle ABC\cong Rt\triangle DEC，$$DF\perp AB，$且$BC = a，$$AC = b，$$AB = c，$
所以$EC = BC = a，$$DC = AC = b，$$DE = AB = c。$
因为$S_{\triangle BCE}+S_{\triangle ACD}=S_{\triangle ABD}-S_{\triangle ABE}，$
所以$\frac{1}{2}a^{2}+\frac{1}{2}b^{2}=\frac{1}{2}c\cdot DF-\frac{1}{2}c\cdot EF=\frac{1}{2}c\cdot(DF - EF)=\frac{1}{2}c\cdot DE=\frac{1}{2}c^{2}。$
所以$a^{2}+b^{2}=c^{2}。$
