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 解：如图，延长$AF$至点$G，$使得$FG = AF，$连接$BG。$

 $\because F$为$BE$的中点，$\therefore EF = BF。$

 在$\triangle AFE$和$\triangle GFB$中，

 $\begin{cases}AF = GF \\\angle AFE=\angle GFB \\EF = BF\end{cases}$

 $\therefore \triangle AFE\cong\triangle GFB(SAS)，$
$\therefore \angle EAF=\angle G，$$AE = GB，$
$\therefore AE// BG，$$\therefore \angle GBA+\angle BAE = 180^{\circ}。$

 $\because \angle BAC+\angle EAD = 180^{\circ}，$
$\therefore \angle DAC+\angle BAE = 180^{\circ}，$
$\therefore \angle GBA=\angle DAC。$

 $\because AD = AE，$$\therefore BG = AD。$

 在$\triangle GBA$和$\triangle DAC$中，

 $\begin{cases}AB = CA \\\angle GBA=\angle DAC \\BG = AD\end{cases}$

 $\therefore \triangle GBA\cong\triangle DAC(SAS)，$
$\therefore AG = CD。$

 $\because AG = AF + FG = 2AF，$
$\therefore CD = 2AF。$

 解：如图，在$BC$上取一点$F，$使得$FB = AB，$连接$DF。$

 $\because BD$平分$\angle ABC，$$\angle ABC = 40^{\circ}，$$\therefore \angle ABD=\angle FBD = 20^{\circ}。$

 在$\triangle ABD$和$\triangle FBD$中，

 $\begin{cases}AB = FB \\\angle ABD=\angle FBD \\BD = BD\end{cases}$

 $\therefore \triangle ABD\cong\triangle FBD(SAS)，$
$\therefore FD = AD，$$\angle BDF=\angle BDA = 180^{\circ}-\angle A-\angle ABD = 60^{\circ}。$

 $\therefore \angle FDC=\angle BDA=\angle EDC = 60^{\circ}。$

 $\because ED = AD，$$\therefore ED = FD。$

 在$\triangle EDC$和$\triangle FDC$中，

 $\begin{cases}ED = FD \\\angle EDC=\angle FDC \\DC = DC\end{cases}$

 $\therefore \triangle EDC\cong\triangle FDC(SAS)，$
$\therefore CE = CF，$$\therefore BC = FB + CF = AB + CE。$

证明： (1) $\because D$是$AB$的中点，$AC = BC，$$\angle ACB = 90^{\circ}，$
$\therefore CD\perp AB，$$\angle ACD=\angle BCD = 45^{\circ}，$
$\angle CAD=\angle CBD = 45^{\circ}，$
$\therefore AC = BC，$$\angle ACE+\angle ECB = 90^{\circ}。$

 $\because BF\perp CE，$$\therefore \angle CBF+\angle ECB = 90^{\circ}，$
$\therefore \angle ACE=\angle CBF。$

 在$\triangle ACE$和$\triangle CBG$中，

 $\begin{cases}\angle ACE=\angle CBF \\AC = BC \\\angle CAE=\angle BCG = 45^{\circ}\end{cases}$

 $\therefore \triangle ACE\cong\triangle CBG(ASA)，$
$\therefore AE = CG。$

 (2) $CM = BE。$

 证明：$\because \angle ACB = 90^{\circ}，$
$\therefore \angle ACH+\angle BCE = 90^{\circ}。$

 $\because AH\perp CE，$
$\therefore \angle ACH+\angle CAH = 90^{\circ}，$$\therefore \angle CAH=\angle BCE。$

 $\because AC = BC，$$\angle AHC=\angle BEC = 90^{\circ}。$

 在$\triangle ACH$和$\triangle CBE$中，

 $\begin{cases}\angle CAH=\angle BCE \\AC = BC \\\angle AHC=\angle BEC\end{cases}$

 $\therefore \triangle ACH\cong\triangle CBE(AAS)，$
$\therefore CH = BE。$

 $\because \angle ACM = 45^{\circ}，$$\angle AHC = 90^{\circ}，$
$\therefore \angle HAC = 45^{\circ}=\angle ACM，$$\therefore AH = CH。$

 $\because \angle ADM=\angle CDH = 90^{\circ}，$$\angle AMD+\angle DAM = 90^{\circ}，$$\angle HCD+\angle DHC = 90^{\circ}，$
$\angle DAM=\angle HCD，$$AD = CD。$

 在$\triangle ADM$和$\triangle CDH$中，

 $\begin{cases}\angle ADM=\angle CDH \\AD = CD \\\angle DAM=\angle DCH\end{cases}$

 $\therefore \triangle ADM\cong\triangle CDH(ASA)，$
$\therefore CM = CH，$$\therefore CM = BE。$