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第115页

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 (1)证明:$\because \overset{\frown}{AD}=\overset{\frown}{CD}，$$\therefore OD\perp AC.$又$\because AB$是$\odot O$的直径，$\therefore \angle ACB = 90^{\circ}，$即$BC\perp AC.$$\therefore OD// BC$

$ (2)解:由 $

 (1)知，$OD\perp AC，$$\therefore AE = CE.$又$\because AC = 10，$$\therefore AE = CE=\frac{1}{2}AC = 5.$设$\odot O$的半径为$R，$则$OA = R，$$OE = R - 4.$在$Rt\triangle AOE$中，$OA^{2}=OE^{2}+AE^{2}，$即$R^{2}=(R - 4)^{2}+5^{2}，$$\therefore R=\frac{41}{8}.$又$\because O，$$E$分别为$AB，$$AC$的中点，$\therefore OE=\frac{1}{2}BC，$$OE// BC.$$\therefore BC = 2OE = 2\times(\frac{41}{8}-4)=\frac{9}{4}$

 (1)证明:如图，连接$OC.$$\because CD$为$\odot O$的切线，点$C$在$\odot O$上，$\therefore \angle OCD = 90^{\circ}，$即$OC\perp CD.$$\because \overset{\frown}{AC}=\overset{\frown}{CE}，$$\therefore OC\perp AE.$$\therefore DC// AE$

 (2)解:如图，连接$OE，$$BE.$$\because EF$垂直平分$OB，$$\therefore OE = BE.$$\because OE = OB，$$\therefore \triangle OEB$为等边三角形.$\therefore \angle BOE = 60^{\circ}.$$\therefore \angle AOE = 180^{\circ}-60^{\circ}=120^{\circ}.$$\because OA = OE，$$\therefore \angle OAE=\angle OEA = 30^{\circ}.$$\because DC// AE，$$\therefore \angle D=\angle OAE = 30^{\circ}.$$\because \angle OCD = 90^{\circ}，$$\therefore OD = 2OC = OA + AD.$$\because OA = OC，$$\therefore OC = AD = 3.$$\therefore AO = OE = OC = 3.$$\therefore$易得$EF=\frac{\sqrt{3}}{2}OE=\frac{3\sqrt{3}}{2}.$$\therefore \triangle OAE$的面积$=\frac{1}{2}AO\cdot FE=\frac{9\sqrt{3}}{4}.$$\because$扇形$OAE$的面积$=\frac{120\pi\times3^{2}}{360}=3\pi，$$\therefore$阴影部分的面积$=$扇形$OAE$的面积$-\triangle OAE$的面积$=3\pi-\frac{9\sqrt{3}}{4}$

 (1)证明:如图，连接$OC.$$\because DC$与$\odot O$相切于点$C，$$\therefore OC\perp CD.$$\because BD\perp CD，$$\therefore OC// BD.$$\therefore \angle OCB = \angle DBC.$$\because OC = OB，$$\therefore \angle OCB = \angle OBC.$$\therefore \angle OBC = \angle DBC.$$\therefore BC$平分$\angle ABD$

 (2)解:如图，过点$C$作$CH\perp AB$于点$H.$$\because AB$为$\odot O$的直径，$\therefore \angle ACB = 90^{\circ}.$$\therefore BC=\sqrt{AB^{2}-AC^{2}}=\sqrt{5^{2}-(2\sqrt{5})^{2}}=\sqrt{5}.$$\because S_{\triangle ABC}=\frac{1}{2}CH\cdot AB=\frac{1}{2}BC\cdot AC，$$\therefore CH=\frac{BC\cdot AC}{AB}=\frac{\sqrt{5}\times2\sqrt{5}}{5}=2.$在$Rt\triangle BCH$中，$BH=\sqrt{BC^{2}-CH^{2}}=\sqrt{(\sqrt{5})^{2}-2^{2}} = 1.$$\because BC$平分$\angle ABD，$$CH\perp BA，$$CD\perp BD，$$\therefore CH = CD.$在$Rt\triangle BCH$和$Rt\triangle BCD$中，$\begin{cases}BC = BC,\\CH = CD,\end{cases}$ $\therefore Rt\triangle BCH\cong Rt\triangle BCD.$$\therefore BH = BD = 1$