解：$S_{四边形ABFD}=S_{长方形DECF}+S_{△ADE}+S_{△ABC}$
$ =b(b-a)+\frac 1 2ab+\frac 1 2ab$
$ ={b}^2$
$S_{四边形ABFD}=S_{△ABD}+S_{△BDF}$
$ =\frac 1 2{c}^2+\frac 1 2(b+a)(b-a)$
$ =\frac 1 2{c}^2+\frac 12{b}^2-\frac 1 2{a}^2$
所以${b}^2=\frac 1 2{c}^2+\frac 12{b}^2-\frac 1 2{a}^2，$
所以${a}^2+{b}^2={c}^2$
