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22.(12分)如图，在$Rt \bigtriangleup ACB$中，$\angle ACB = 90^{\circ}$，$AC = BC$，点$E$为射线$CB$上一动点，连接$AE$，作$AF\bot AE$，且$AF = AE$.
(1)如图1，连接$EF$，过点$F$作$FD\bot AC$于点$D$，求证：$EC + CD = DF$.
(2)如图2，连接$BF$交$AC$于点$G$，若$\frac{AG}{CG} = 3$，求证：点$E$为$BC$的中点.
(3)当点$E$在射线$CB$上时，连接$BF$，交直线$AC$于点$G.$若$\frac{BC}{BE} = \frac{4}{3}$，则$\frac{AG}{CG} =$
(1)如图1，连接$EF$，过点$F$作$FD\bot AC$于点$D$，求证：$EC + CD = DF$.
(2)如图2，连接$BF$交$AC$于点$G$，若$\frac{AG}{CG} = 3$，求证：点$E$为$BC$的中点.
(3)当点$E$在射线$CB$上时，连接$BF$，交直线$AC$于点$G.$若$\frac{BC}{BE} = \frac{4}{3}$，则$\frac{AG}{CG} =$
$\frac{11}{3}$或$\frac{5}{3}$
.
答案：
22.
(1)证明：$\because AF\perp AE,\therefore \angle FAE=90^{\circ}$. $\therefore \angle FAD+\angle CAE=90^{\circ}$. $\because FD\perp AC,\therefore \angle FDA=90^{\circ}$. $\therefore \angle FAD+\angle DFA=90^{\circ}$. $\therefore \angle CAE=\angle DFA$. 在$\triangle ADF$和$\triangle ECA$中，$\begin{cases}\angle ADF=\angle ECA,\\\angle DFA=\angle CAE,\\AF=EA,\end{cases}$ $\therefore \triangle ADF\cong\triangle ECA(AAS).\therefore AD=EC$，$FD=AC$. $\therefore EC+CD=AD+CD=AC=FD$. $\therefore EC+CD=DF$.
(2)证明：如图1，过点$F$作$FD\perp AC$于点$D$. $\therefore \angle FDG=\angle C=90^{\circ}$. 由
(1)得，$\triangle ADF\cong\triangle ECA$，$\therefore FD=AC=BC$，$AD=EC$. 在$\triangle FDG$和$\triangle BCG$中，$\begin{cases}\angle FGD=\angle BGC,\\\angle FDG=\angle C,\\FD=BC,\end{cases}$ $\therefore \triangle FDG\cong\triangle BCG(AAS).\therefore DG=CG$. $\because \frac{AG}{CG}=3$，$\therefore AG=3CG$. $\therefore AD=AG - DG=3CG - CG=2CG$，$AC=AG + CG=4CG$. $\therefore EC=AD=2CG$，$BC=AC=4CG$. $\therefore BC=2EC$，$\therefore EB=EC$. $\therefore$点$E$为$BC$的中点.
(3)如图2，当点$E$在线段$CB$的延长线上时，过点$F$作$FD\perp AG$交$AG$的延长线于点$D$. $\because \frac{BC}{BE}=\frac{4}{3}$，$\therefore$设$BC=4x$，则$BE=3x$. $\therefore AC=BC=4x$，$CE=BC + BE=7x$. 由
(1)
(2)得，$\triangle ADF\cong\triangle ECA$，$\triangle GDF\cong\triangle GCB$. $\therefore AD=EC=7x$. $\therefore GD=GC=\frac{1}{2}(AD - AC)=\frac{3}{2}x$. $\therefore AG=AC + CG=\frac{11}{2}x$. $\therefore \frac{AG}{CG}=\frac{\frac{11}{2}x}{\frac{3}{2}x}=\frac{11}{3}$. 同理，当点$E$在线段$BC$上时，$\frac{AG}{CG}=\frac{5}{3}$. 综上所述，$\frac{AG}{CG}=\frac{11}{3}$或$\frac{5}{3}$.
(1)证明：$\because AF\perp AE,\therefore \angle FAE=90^{\circ}$. $\therefore \angle FAD+\angle CAE=90^{\circ}$. $\because FD\perp AC,\therefore \angle FDA=90^{\circ}$. $\therefore \angle FAD+\angle DFA=90^{\circ}$. $\therefore \angle CAE=\angle DFA$. 在$\triangle ADF$和$\triangle ECA$中，$\begin{cases}\angle ADF=\angle ECA,\\\angle DFA=\angle CAE,\\AF=EA,\end{cases}$ $\therefore \triangle ADF\cong\triangle ECA(AAS).\therefore AD=EC$，$FD=AC$. $\therefore EC+CD=AD+CD=AC=FD$. $\therefore EC+CD=DF$.
(2)证明：如图1，过点$F$作$FD\perp AC$于点$D$. $\therefore \angle FDG=\angle C=90^{\circ}$. 由
(1)得，$\triangle ADF\cong\triangle ECA$，$\therefore FD=AC=BC$，$AD=EC$. 在$\triangle FDG$和$\triangle BCG$中，$\begin{cases}\angle FGD=\angle BGC,\\\angle FDG=\angle C,\\FD=BC,\end{cases}$ $\therefore \triangle FDG\cong\triangle BCG(AAS).\therefore DG=CG$. $\because \frac{AG}{CG}=3$，$\therefore AG=3CG$. $\therefore AD=AG - DG=3CG - CG=2CG$，$AC=AG + CG=4CG$. $\therefore EC=AD=2CG$，$BC=AC=4CG$. $\therefore BC=2EC$，$\therefore EB=EC$. $\therefore$点$E$为$BC$的中点.
(3)如图2，当点$E$在线段$CB$的延长线上时，过点$F$作$FD\perp AG$交$AG$的延长线于点$D$. $\because \frac{BC}{BE}=\frac{4}{3}$，$\therefore$设$BC=4x$，则$BE=3x$. $\therefore AC=BC=4x$，$CE=BC + BE=7x$. 由
(1)
(2)得，$\triangle ADF\cong\triangle ECA$，$\triangle GDF\cong\triangle GCB$. $\therefore AD=EC=7x$. $\therefore GD=GC=\frac{1}{2}(AD - AC)=\frac{3}{2}x$. $\therefore AG=AC + CG=\frac{11}{2}x$. $\therefore \frac{AG}{CG}=\frac{\frac{11}{2}x}{\frac{3}{2}x}=\frac{11}{3}$. 同理，当点$E$在线段$BC$上时，$\frac{AG}{CG}=\frac{5}{3}$. 综上所述，$\frac{AG}{CG}=\frac{11}{3}$或$\frac{5}{3}$.
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