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第45页

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 (1)证明：$\because CE// BD,EB// AC，$

 $\therefore$四边形$OCEB$是平行四边形.

 $\because$四边形$ABCD$是菱形，$\therefore AC\perp BD，$即$\angle BOC = 90^{\circ}.$

 $\therefore$平行四边形$OCEB$是矩形.

 $\therefore OE = CB.$

 (2)解：$\because OC:OB = 1:2，$$\therefore$设$OC = x，$则$OB = 2x.$

由

 (1)知$CB = OE = 2\sqrt{5}.$

 在$Rt\triangle OBC$中，由勾股定理，得$x^{2}+(2x)^{2}=(2\sqrt{5})^{2}，$

 解得$x=\pm2$（负值舍去）.

 $\therefore OC = 2，$$OB = 4.$

 $\therefore S_{菱形ABCD}=4S_{\triangle OBC}=4\times\frac{1}{2}\times2\times4 = 16.$

 (1)证明：由平移的性质，得$AE// DF，$$AE = DF，$

 $\therefore$四边形$AEFD$是平行四边形.

 $\because AE\perp BC，$$\therefore\angle AEF = 90^{\circ}.$

 $\therefore$平行四边形$AEFD$是矩形.

 (2)解：由

 (1)，得四边形$AEFD$是矩形，$\therefore\angle DFE = 90^{\circ}.$

 $\therefore BF=\sqrt{BD^{2}-DF^{2}}=\sqrt{(4\sqrt{5})^{2}-4^{2}} = 8.$

 $\because$四边形$ABCD$是菱形，

 $\therefore OA = OC，$$OB = OD=\frac{1}{2}BD = 2\sqrt{5}，$$AC\perp BD，$$AB = BC = CD.$

 设$AB = BC = CD = x，$则$CF = 8 - x，$

 在$Rt\triangle CDF$中，由勾股定理得$(8 - x)^{2}+4^{2}=x^{2}，$

 解得$x = 5，$$\therefore AB = 5.$

 在$Rt\triangle AOB$中，由勾股定理，得$OA=\sqrt{AB^{2}-OB^{2}}=\sqrt{5^{2}-(2\sqrt{5})^{2}}=\sqrt{5}，$$\therefore AC = 2OA = 2\sqrt{5}.$

 (1)证明：$\because$四边形$EFGH$是矩形，

 $\therefore EH = FG，$$EH// FG.$ $\therefore\angle GFH=\angle EHF.$

 $\because\angle BFG = 180^{\circ}-\angle GFH，$$\angle DHE = 180^{\circ}-\angle EHF，$

 $\therefore\angle BFG=\angle DHE.$

 $\because$四边形$ABCD$是菱形，$\therefore AD// BC.$

 $\therefore\angle GBF=\angle EDH.$ $\therefore\triangle BGF\cong\triangle DEH(AAS).$

 $\therefore BG = DE.$

 (2)解：如答图，连接$EG.$

 $\because$四边形$EFGH$是矩形，$\therefore FH = EG.$

 $\because$四边形$ABCD$是菱形，$\therefore AD = BC，$$AD// BC.$

 $\because E$为$AD$的中点，$\therefore AE = ED.$

 $\because BG = DE，$$\therefore AE = BG，$$\because AE// BG，$

 $\therefore$四边形$ABGE$是平行四边形. $\therefore AB = EG.$

 $\because EG = FH = 2，$$\therefore AB = 2.$ $\therefore$菱形$ABCD$的周长$= 8.$