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第81页

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 （1）证明:$\because$ $\triangle ABC$绕点$A$按逆时针方向旋转$100^{\circ}，$

 $\therefore$ $\angle BAC=\angle DAE = 40^{\circ}，$$\angle BAD=\angle CAE = 100^{\circ}。$又$\because$ $AB = AC，$$\therefore$ $AB = AC = AD = AE。$

 在$\triangle ABD$和$\triangle ACE$中，

 $\begin{cases}AB = AC\\\angle BAD=\angle CAE\\AD = AE\end{cases}，$$\therefore$ $\triangle ABD\cong\triangle ACE。$

 （2）解:四边形$ABFE$是菱形

 理由：$\because$ $\angle BAD=\angle CAE = 100^{\circ}，$$AB = AC = AD = AE，$

 $\therefore$ $\angle ABD=\angle ADB=\angle ACE=\angle AEC = 40^{\circ}。$$\because$ $\angle BAE=\angle BAD+\angle DAE = 140^{\circ}，$

 $\therefore$ $\angle BFE = 360^{\circ}-\angle BAE-\angle ABD-\angle AEC = 140^{\circ}。$$\therefore$ $\angle BAE=\angle BFE。$$\therefore$ 四边形$ABFE$是平行四边形。$\because$ $AB = AE，$$\therefore$ 四边形$ABFE$是菱形。

解: （1）如图，$\triangle A_{1}B_{1}C_{1}$即为所求作。

 （2）如图，$\triangle A_{1}C_{1}C_{2}$的面积$=4\times8-\frac{1}{2}\times3\times2-\frac{1}{2}\times2\times8-\frac{1}{2}\times4\times5 = 11$

解:

 （1）$\because$ 四边形$ABCD$是正方形，$\therefore$ $\angle BAD = 90^{\circ}，$$AB = AD。$

 $\because$ 将线段$AB$按顺时针方向旋转$\alpha(0^{\circ}\lt\alpha\lt90^{\circ})，$得到线段$AE，$$\therefore$ $\angle EAB=\alpha，$$AB = AE。$

 $\therefore$ $AE = AD，$$\angle EAD = 90^{\circ}+\alpha。$

 $\therefore$ $\angle AED=\frac{180^{\circ}-(90^{\circ}+\alpha)}{2}=45^{\circ}-\frac{1}{2}\alpha。$$\because$ $AE = AB，$$\angle EAB=\alpha，$

 $\therefore$ $\angle AEB=\frac{180^{\circ}-\alpha}{2}=90^{\circ}-\frac{1}{2}\alpha。$$\therefore$ $\angle DEB=\angle AEB-\angle AED=(90^{\circ}-\frac{1}{2}\alpha)-(45^{\circ}-\frac{1}{2}\alpha)=45^{\circ}。$

 （2）补全图形如图所示，$DE = \sqrt{2}CF。$

 如图，过点$C$作$CG\perp CF，$交$FD$的延长线于点$G。$$\because$ $BF\perp DE，$$\therefore$ $\angle EFB=\angle BFD = 90^{\circ}。$

 $\therefore$ $\angle BFC+\angle CFD = 90^{\circ}。$$\because$ $CG\perp CF，$$\therefore$ $\angle FCG = 90^{\circ}。$

 $\therefore$ $\angle CFD+\angle G = 90^{\circ}。$$\therefore$ $\angle BFC=\angle G。$

 在正方形$ABCD$中，$BC = CD，$$\angle BCD = 90^{\circ}。$$\because$ $\angle BCD=\angle FCG = 90^{\circ}，$$\therefore$ $\angle BCF=\angle DCG。$

 $\because$ $BC = CD，$$\therefore$ $\triangle BCF\cong\triangle DCG。$$\therefore$ $BF = DG，$$CF = CG。$

 $\therefore$ $\triangle FCG$是等腰直角三角形。$\therefore$ 易得$FG=\sqrt{2}CF。$

 由（1），知$\angle DEB = 45^{\circ}，$$\therefore$ $\triangle BEF$是等腰直角三角形。$\therefore$ $EF = BF。$

 $\therefore$ $EF = DG。$$\therefore$ $EF + FD = DG+FD，$即$DE = FG。$$\therefore$ $DE=\sqrt{2}CF。$