11. 如图,$B, D, C$ 三点在一条直线上,$\angle A D B=\angle A D C=90^{\circ}, B D=D E, \angle D A C=45^{\circ}$.
(1) 试判断线段 $A B, C E$ 的关系,并说明理由;
(2) 若 $B D=a, A D=b, A B=c$,请利用此图的面积证明勾股定理.
答案：(1) 解:$AB = CE$,$AB\perp CE$.
理由:延长$CE$交$AB$于点$F$,如答图.
$\because\angle ADC = 90^{\circ}$,$\angle DAC = 45^{\circ}$,
$\therefore\angle ACD = \angle DAC = 45^{\circ}$,$\therefore AD = CD$.
在$\triangle ADB$和$\triangle CDE$中,$\begin{cases}AD = CD\\\angle ADB = \angle CDE\\DB = DE\end{cases}$
$\therefore\triangle ADB\cong\triangle CDE(\mathrm{SAS})$,
$\therefore AB = CE$,$\angle BAD = \angle DCE$.
$\because\angle BAD + \angle ABD = 90^{\circ}$,
$\therefore\angle DCE + \angle ABD = 90^{\circ}$,$\therefore\angle BFC = 90^{\circ}$,
$\therefore AB\perp CE$.
(2) 证明:设$EF = x$,
$\because S_{\triangle ABC}=S_{\triangle ABE}+S_{\triangle BDE}+S_{\triangle ACD}$,
$\therefore\frac{1}{2}AB\cdot CF=\frac{1}{2}AB\cdot EF+\frac{1}{2}BD\cdot DE+\frac{1}{2}DC\cdot AD$.
$\because BD = a$,$AB = c$,$AD = b$,$\triangle ADB\cong\triangle CDE$,
$\therefore AB = CE = c$,$BD = DE = a$,$AD = CD = b$,
$\therefore\frac{1}{2}c(c + x)=\frac{1}{2}cx+\frac{1}{2}a^{2}+\frac{1}{2}b^{2}$,
即$c^{2}+cx = cx + a^{2}+b^{2}$,
$\therefore a^{2}+b^{2}=c^{2}$.
