解：?$S_{四邊形ABFD}=S_{長(zhǎng)方形DECF}+S_{△ADE}+S_{△ABC}$?
?$ =b(b-a)+\frac 1 2ab+\frac 1 2ab$?
?$ =^2$?
?$S_{四邊形ABFD}=S_{△ABD}+S_{△BDF}$?
?$ =\frac 1 2{c}^2+\frac 1 2(b+a)(b-a)$?
?$ =\frac 1 2{c}^2+\frac 12^2-\frac 1 2{a}^2$?
所以?$^2=\frac 1 2{c}^2+\frac 12^2-\frac 1 2{a}^2，$?
所以?${a}^2+^2={c}^2$?
