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第25页

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$(1,4)$

 （1）$\triangle ACD\cong\triangle CBE$

 证明：$\because \angle ACB = 90^{\circ}，$
$\therefore \angle ACD + \angle ECB = 90^{\circ}.$

 $\because AD\perp CE，$$BE\perp CE，$
$\therefore \angle ADC = \angle CEB = 90^{\circ}.$

 $\therefore \angle ACD + \angle DAC = 90^{\circ}，$
$\therefore \angle ECB = \angle DAC.$

 在$\triangle ACD$和$\triangle CBE$中，

 $\begin{cases}\angle ADC = \angle CEB \\\angle DAC = \angle ECB \\AC = CB\end{cases}$

 $\therefore \triangle ACD\cong\triangle CBE$（AAS）

 （2）$\because \triangle ACD\cong\triangle CBE，$
$\therefore CD = BE = 3，$$AD = CE.$

 又$\because CE = CD + DE = 3 + 5 = 8，$
$\therefore AD = 8$

 （1）证明：在$\triangle ABE$和$\triangle ACD$中，

 $\begin{cases}\angle B = \angle C \\AB = AC \\\angle BAE = \angle CAD\end{cases}$

 $\therefore \triangle ABE\cong\triangle ACD$（ASA）.

 $\therefore AE = AD.$

 $\therefore AB - AD = AC - AE，$即$BD = CE.$

 （2）证明：在$\triangle BDF$和$\triangle CEF$中，

 $\begin{cases}\angle B = \angle C \\\angle BFD = \angle CFE \\BD = CE\end{cases}$

 $\therefore \triangle BDF\cong\triangle CEF$（AAS）.

 （3）证明：由（2），知$\triangle BDF\cong\triangle CEF，$$\therefore BF = CF.$

 在$\triangle ABF$和$\triangle ACF$中，

 $\begin{cases}AB = AC \\\angle B = \angle C \\BF = CF\end{cases}$

 $\therefore \triangle ABF\cong\triangle ACF$（SAS）.

 $\therefore \angle BAF = \angle CAF，$即$AF$平分$\angle BAC.$