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第93页

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 （1）证明：$\because DA$平分$\angle BDF，$$\therefore \angle ADF=\angle ADB.$$\because \angle ABC+\angle ADC = 180^{\circ}，$$\angle ADC+\angle ADF = 180^{\circ}，$$\therefore \angle ADF=\angle ABC.$$\therefore \angle ADB=\angle ABC.$$\because \angle ACB=\angle ADB，$$\therefore \angle ABC=\angle ACB.$$\therefore AB = AC$

 （2）解：如图，过点$A$作$AG\perp BD，$垂足为$G.$$\because DA$平分$\angle BDF，$$AE\perp CF，$$AG\perp BD，$$\therefore AG = AE，$$\angle AGB=\angle AGD=\angle AEC = 90^{\circ}.$ 在$Rt\triangle AED$和$Rt\triangle AGD$中，$\begin{cases}AD = AD\\AE = AG\end{cases}，$$\therefore Rt\triangle AED\cong Rt\triangle AGD.$$\therefore DE = DG = 2.$ 在$Rt\triangle AEC$和$Rt\triangle AGB$中，$\begin{cases}AC = AB\\AE = AG\end{cases}，$$\therefore Rt\triangle AEC\cong Rt\triangle AGB.$$\therefore CE = BG.$$\because BD = 11，$$\therefore BG = BD - DG = 11 - 2 = 9.$$\therefore CE = BG = 9.$$\therefore CD = CE - DE = 9 - 2 = 7$

 （1）解：$\because \angle DAB = 90^{\circ}，$$\therefore BD$为$\odot O$的直径，即$BD = 12.$$\because \overset{\frown}{AD}=\overset{\frown}{AB}，$$\therefore AD = AB.$$\therefore \triangle ABD$为等腰直角三角形.$\therefore$易得$AB=\frac{\sqrt{2}}{2}BD = 6\sqrt{2}$

 （2）解：连接$BD，$过点$B$作$BH\perp AC$于点$H.$$\because \angle DAB = 90^{\circ}，$$\therefore BD$为$\odot O$的直径，且$BD=\sqrt{AD^{2}+AB^{2}}=\sqrt{5^{2}+3^{2}}=\sqrt{34}.$$\therefore \angle BCD = 90^{\circ}.$$\because AC$平分$\angle DAB，$$\therefore \angle BAC=\angle DAC = 45^{\circ}.$$\therefore BC = DC.$$\therefore \triangle BCD$为等腰直角三角形.$\therefore$易得$BC=\frac{\sqrt{2}}{2}BD=\frac{\sqrt{2}}{2}\times\sqrt{34}=\sqrt{17}.$ 在$Rt\triangle ABH$中，$\because \angle BAC = 45^{\circ}，$$\therefore$易得$AH = BH=\frac{\sqrt{2}}{2}AB=\frac{3\sqrt{2}}{2}.$ 在$Rt\triangle BCH$中，$CH=\sqrt{BC^{2}-BH^{2}}=\sqrt{(\sqrt{17})^{2}-(\frac{3\sqrt{2}}{2})^{2}}=\frac{5\sqrt{2}}{2}，$$\therefore AC = AH + CH=\frac{3\sqrt{2}}{2}+\frac{5\sqrt{2}}{2}=4\sqrt{2}$