$证明:(1)∵AC⊥BD,∠CAD=45°$
$∴AC=DC,∠ACB=∠DCE= 90°$
$在Rt△ABC和Rt△DEC中$
$\begin{cases}{AB=DE} \\ {AC=DC} \end{cases}$
$∴Rt△ABC≌Rt△DEC(\mathrm {HL})$
$∴∠BAC=∠EDC$
$∵∠EDC+∠CED=90°,∠CED=∠AEF$
$∴∠AEF+∠BAC= 90°$
$∴∠AFE=90°$
$∴DF⊥AB$
$(2)∵S_{△BCE} +S_{△ACD}= S_{△ABD}-S_{△ABE}$
$∴\frac{1}{2}a²+\frac{1}{2}b²=\frac{1}{2}×c×DF-\frac{1}{2}×c×EF=\frac{1}{2}×c×(DF-EF)=\frac{1}{2}×c×DE=\frac{1}{2}c²$
$∴a²+b²=c²$