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第56页

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证明: (1)$\because \angle ABC = \angle ADC = 90^{\circ}$

$M，$$N$分别是$AC，$$BD$的中点，$\therefore$在$Rt\triangle ABC$中，$BM=\frac{1}{2}AC，$在$Rt\triangle ACD$中，$DM=\frac{1}{2}AC，$$\therefore BM = DM。$又$\because N$是$BD$的中点，$\therefore MN\perp BD。$

 (2)$\triangle MBD$是等腰直角三角形。理由：$\because M$是$AC$的中点，$\therefore AM=\frac{1}{2}AC = BM，$$\therefore \angle BAM = \angle ABM，$$\therefore \angle BMC = 2\angle BAM。$同理可得$\angle DMC = 2\angle DAM。$又$\because \angle BAD = 45^{\circ}，$$\therefore \angle BMD = \angle BMC + \angle DMC = 2(\angle BAM + \angle DAM)=2\angle BAD = 90^{\circ}。$又$\because BM = DM，$$\therefore \triangle MBD$是等腰直角三角形。

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证明: 由题图③，得$CE$是$\triangle AEF$的中线，$\therefore S_{\triangle ACE}=S_{\triangle EFC}。$$\because \angle ACB = 90^{\circ}，$$\therefore \angle BCF = 90^{\circ}。$$\because \triangle CBE\cong\triangle CAD，$$\therefore CE = CD，$$\angle ECB = \angle DCA，$$\therefore 90^{\circ}-\angle ECB = 90^{\circ}-\angle DCA，$即$\angle ECF = \angle DCB。$又$\because AC = BC，$$AC = CF，$$\therefore CF = CB。$在$\triangle ECF$和$\triangle DCB$中，$\begin{cases}CF = CB \\ \angle ECF = \angle DCB \\ CE = CD\end{cases}，$$\therefore \triangle ECF\cong\triangle DCB(SAS)。$$\therefore S_{\triangle ECF}=S_{\triangle DCB}，$$\therefore S_{\triangle CAE}=S_{\triangle CDB}。$

证明: (1)$\because DE\perp AB，$$\angle ACB = 90^{\circ}，$$\therefore \angle AED = \angle AEF = \angle ACB = 90^{\circ}。$在$Rt\triangle ACF$与$Rt\triangle AEF$中，$\begin{cases}AC = AE \\ AF = AF\end{cases}，$$\therefore Rt\triangle ACF\cong Rt\triangle AEF(HL)，$$\therefore CF = EF。$在$Rt\triangle ADE$与$Rt\triangle ABC$中，$\begin{cases}AD = AB \\ AE = AC\end{cases}，$$\therefore Rt\triangle ADE\cong Rt\triangle ABC(HL)，$$\therefore DE = BC。$$\because DF = DE + EF，$$\therefore DF = BC + CF。$

 (2)$BC = CF + DF。$证明如下：连接$AF，$在$Rt\triangle ABC$与$Rt\triangle ADE$中，$\begin{cases}AB = AD \\ AC = AE\end{cases}，$$\therefore Rt\triangle ABC\cong Rt\triangle ADE(HL)，$$\therefore BC = DE。$$\because \angle ACB = 90^{\circ}，$$\therefore \angle ACF = 90^{\circ}=\angle AED。$在$Rt\triangle ACF$与$Rt\triangle AEF$中，$\begin{cases}AC = AE \\ AF = AF\end{cases}，$$\therefore Rt\triangle ACF\cong Rt\triangle AEF(HL)，$$\therefore CF = EF。$$\because DE = EF + DF，$$\therefore BC = CF + DF。$