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（1）证明：因为$AG\perp AE，$所以$\angle EAG = 90^{\circ}，$所以$\angle EDG = \angle EAG = 90^{\circ}。$因为四边形$ABCD$是正方形，所以$AD = CD，$$\angle ADC = 90^{\circ}。$因为$\angle ADF+\angle FDC = \angle CDG+\angle FDC = 90^{\circ}，$所以$\angle ADF = \angle CDG。$在$\triangle ADF$和$\triangle CDG$中，$\begin{cases}\angle FAD = \angle GCD\\AD = CD\\\angle ADF = \angle CDG\end{cases}，$所以$\triangle ADF\cong\triangle CDG$（$ASA$）。

（2）解：过点$D$作$DH\perp AG$于点$H，$连接$OA，$$OD。$因为四边形$ABCD$是正方形，所以$AB = AD = 2，$$\angle BAD = 90^{\circ}，$$\angle AOD = 90^{\circ}，$所以$\angle AGD=\frac{1}{2}\angle AOD = 45^{\circ}。$因为$DH\perp AG，$所以$\angle DHG = 90^{\circ}，$所以$\triangle HDG$是等腰直角三角形，所以$HG = DH。$因为$\angle DAG+\angle BAG = \angle BAE+\angle BAG = 90^{\circ}，$所以$\angle DAG = \angle BAE = 30^{\circ}。$在$Rt\triangle ADH$中，$\angle DAG = 30^{\circ}，$所以$DH=\frac{1}{2}AD = 1，$$AH = \sqrt{AD^{2}-DH^{2}}=\sqrt{4 - 1}=\sqrt{3}，$$HG = DH = 1，$所以$AG = AH + HG=\sqrt{3}+1，$$DG = \sqrt{DH^{2}+HG^{2}}=\sqrt{2}。$由（1）得$\triangle ADF\cong\triangle CDG，$所以$DF = DG=\sqrt{2}。$$S_{\triangle ADF}=S_{\triangle ADG}-S_{\triangle DFG}=\frac{1}{2}\times(\sqrt{3}+1)\times1-\frac{1}{2}\times\sqrt{2}\times\sqrt{2}=\frac{\sqrt{3}-1}{2}。$因为$OA = OD，$$\angle AOD = 90^{\circ}，$所以$\triangle AOD$是等腰直角三角形，$OA=\frac{\sqrt{2}}{2}AD=\sqrt{2}。$$S_{弓形AD}=S_{扇形OAD}-S_{\triangle AOD}=\frac{90\pi\times(\sqrt{2})^{2}}{360}-\frac{1}{2}\times\sqrt{2}\times\sqrt{2}=\frac{\pi}{2}-1。$所以$S_{涂色部分}=S_{\triangle ADF}+S_{弓形AD}=\frac{\sqrt{3}-1}{2}+\frac{\pi}{2}-1=\frac{\sqrt{3}+\pi - 3}{2}。$

（1）解:因为$\angle BAE = \angle CAD，$所以$\angle BAE+\angle BAD = \angle CAD+\angle BAD，$即$\angle EAD = \angle BAC。$又因为$\angle ADE = \angle ACB，$$AD = AC，$所以$\triangle ADE\cong\triangle ACB$（$ASA$），所以$AE = AB。$因为$AB = 8，$所以$AE = 8。$

（2）证明：连接$BO$并延长，交$\odot O$于点$F，$连接$AF。$因为$BF$是$\odot O$的直径，所以$\angle BAF = 90^{\circ}，$所以$\angle AFB+\angle ABF = 90^{\circ}。$因为$\angle AFB = \angle ACB，$所以$\angle ACB+\angle ABF = 90^{\circ}。$在$\triangle ADC$中，$AD = AC，$所以$\angle ACB = \angle ADC，$所以$2\angle ACB+\angle CAD = 180^{\circ}。$由（1）知$AE = AB，$所以$\angle AEB = \angle ABE，$所以$2\angle ABE+\angle BAE = 180^{\circ}。$因为$\angle BAE = \angle CAD，$所以$\angle ACB = \angle ABE，$所以$\angle ABE+\angle ABF = 90^{\circ}，$即$\angle OBE = 90^{\circ}。$因为$OB$为$\odot O$的半径，所以$EB$是$\odot O$的切线。

（1）证明：连接$EF。$因为$AE$平分$\angle BAC，$所以$\angle BAE = \angle CAE。$因为$FE = FA，$所以$\angle FEA = \angle BAE，$所以$\angle CAE = \angle FEA，$所以$EF// AC，$所以$\angle FEB = \angle C = 90^{\circ}，$所以$EF\perp BC。$又因为$EF$是$\odot F$的半径，所以$BC$是$\odot F$的切线。

（2）解：连接$FD。$因为点$A$的坐标为$(0,-1)，$点$D$的坐标为$(2,0)，$所以$OA = 1，$$OD = 2。$设$\odot F$的半径为$R，$则$OF = R - 1。$在$Rt\triangle FOD$中，根据勾股定理$OF^{2}+OD^{2}=FD^{2}，$即$(R - 1)^{2}+2^{2}=R^{2}，$展开得$R^{2}-2R + 1+4 = R^{2}，$移项化简得$2R = 5，$解得$R = 2.5。$所以$\odot F$的半径为$2.5。$

（3）$AG = AD + 2CD。$证明：过点$E$作$EM\perp AG，$垂足为$M。$因为$\angle C = 90^{\circ}，$所以$EC\perp AC。$又因为$AE$平分$\angle BAC，$$EM\perp AG，$所以$EM = EC。$在$Rt\triangle AEM$和$Rt\triangle AEC$中，$\begin{cases}AE = AE\\EM = EC\end{cases}，$所以$Rt\triangle AEM\cong Rt\triangle AEC$（$HL$），所以$AM = AC，$所以$AG - MG = AD + CD。$连接$GE，$$ED。$因为$\angle BAE = \angle CAE，$所以$\overset{\frown}{EG}=\overset{\frown}{ED}，$所以$EG = ED。$又因为$EM = EC，$所以$Rt\triangle GEM\cong Rt\triangle DEC$（$HL$），所以$MG = CD，$所以$AG - CD = AD + CD，$即$AG = AD + 2CD。$