$证明：(1)∵A C \perp B D，\angle CA D=45°$
$∴A C=D C，\angle A C B=\angle D C E= 90°\ $
$在 Rt \triangle A B C 与 Rt \triangle D E C 中$
$\begin{cases}{A C=D C}\\{A B=D E}\end{cases}$
$∴Rt \triangle A B C≌\ Rt \triangle D E C(\mathrm{HL})，∴\angle BA C=\angle E D C\ $
$∵\angle E D C+\angle C E D=90°，\angle C E D=\angle A E F$
$∴\angle A E F+\angle BA C=90°$
$∴\angle A F E=90°，∴D F \perp A B\ $
$(2)由 Rt \triangle A B C ≌ \mathrm{Rt} \triangle D E C$
$得 B C=E C=a，A B=D E=c，A C= C D=b\ $
$∵S_{\triangle B C E}+S_{\triangle A C D}=S_{\triangle A B D}-S_{\triangle A B E}$
$∴\frac{1}{2}\ \mathrm {a}^{2}+\frac{1}{2}\ \mathrm {b}^{2}= \frac{1}{2} ·c ·D F-\frac{1}{2} ·c ·E F$
$=\frac{1}{2} ·c ·(D F-E F)=\frac{1}{2} ·c ·D E= \frac{1}{2}\ \mathrm {c}^{2}$
$∴a^{2}+b^{2}=c^{2}\ $
