零五网 › 全部参考答案› 启东中学作业本 › 启东中学作业本八年级数学江苏版宿迁专版 › 第57页

第57页

信息发布者：

 证明：如答图，取$BE$的中点$H，$连接$CH,FH.$

 $\because F$是$AE$的中点，$H$是$BE$的中点，$\therefore FH$是$\triangle ABE$的中位线.$\therefore FH// AB，$$FH=\frac{1}{2}AB.$

 又$\because E$是$CD$的中点，$\therefore EC=\frac{1}{2}DC.$

 $\because$四边形$ABCD$是平行四边形，$\therefore AB// DC，$$AB = DC.$

 $\therefore FH = EC，$$FH// EC.$

 $\therefore$四边形$EFHC$是平行四边形.$\therefore GF = GC.$

 (1)证明：由题意知$\angle BAD=\angle CAD，$$\angle AEB=\angle AEF = 90^{\circ}，$

 $\because AE = AE，$$\therefore\triangle ABE\cong\triangle AFE，$$\therefore BE = EF.$

 (2)如答图，取$BC$的中点$M，$连接$EM，$

由

 (1)知$\triangle ABE\cong\triangle AFE，$$\therefore BE = EF，$$AB = AF，$

 $\therefore ME=\frac{1}{2}CF，$$ME// AF，$$\therefore\angle EMC=\angle ACD.$

 $\because AD = AC，$$\therefore\angle ACD=\angle ADC，$

 又$\because\angle ADC=\angle MDE，$$\therefore\angle MDE=\angle EMD，$

 $\therefore DE = ME，$$\therefore AB - AC = CF = 2ME = 2DE.$

 解：如答图，延长$BD,CA$交于点$E，$

 易证$AE = AB，$$BD = ED，$

 $\because BM = CM，$
$\therefore DM=\frac{1}{2}CE=\frac{1}{2}(AB + AC)=15.$

 (1)证明：由旋转的性质得$DM = DE，$$\angle MDE = 2\alpha，$

 $\because\angle C=\alpha，$$\therefore\angle DEC=\angle MDE-\angle C=\alpha，$

 $\therefore\angle C=\angle DEC，$

 $\therefore DE = DC，$$\therefore DM = DC，$

 $\therefore D$是$MC$的中点.

 (2)解：$\angle AEF = 90^{\circ}.$证明如下：

 如答图，延长$FE$到点$H$使$EH = FE，$连接$CH,AH.$

 $\because DF = DC，$$\therefore DE$是$\triangle FCH$的中位线，

 $\therefore DE// CH，$$CH = 2DE.$

 由旋转的性质得$DM = DE，$$\angle MDE = 2\alpha，$

 $\therefore\angle FCH = 2\alpha.$

 $\because\angle B=\angle ACB=\alpha，$

 $\therefore\angle ACH=\alpha，$$\triangle ABC$是等腰三角形，

 $\therefore\angle B=\angle ACH，$$AB = AC，$

 设$DM = DE = m，$$CD = n，$则$CH = 2m，$$CM = m + n，$

 $\therefore DF = CD = n，$$\therefore FM = DF - DM = n - m.$

 $\because AM\perp BC，$$\therefore BM = CM = m + n，$

 $\therefore BF = BM - FM = m + n-(n - m)=2m，$

 $\therefore CH = BF.$

 在$\triangle ABF$和$\triangle ACH$中，

$\begin{cases}AB = AC,\\\angle B=\angle ACH,\\BF = CH,\end{cases}$

 $\therefore\triangle ABF\cong\triangle ACH(SAS)，$$\therefore AF = AH.$

 $\because FE = EH，$$\therefore AE\perp FH，$即$\angle AEF = 90^{\circ}.$