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第24页

信息发布者：

$AB = DE$

$\angle ACB = \angle DFE$

$\angle B = \angle E$

 证明：$\because AC = BD，$

 $\therefore AC + BC = BD + BC，$即$AB = CD.$

 在$\triangle ABF$和$\triangle CDE$中，

 $\begin{cases}\angle A = \angle DCE \\\angle F = \angle E \\AB = CD\end{cases}$

 $\therefore \triangle ABF\cong\triangle CDE$（AAS）

 证明：$\because AD\perp BC，$$\therefore \angle ADB = 90^{\circ}，$$\therefore \angle B + \angle EAF = 90^{\circ}.$

 $\because CE\perp AB，$$\therefore \angle AEF = \angle CEB = 90^{\circ}，$$\therefore \angle B + \angle ECB = 90^{\circ}.$

 $\therefore \angle EAF = \angle ECB.$

 在$\triangle AEF$和$\triangle CEB$中，

 $\begin{cases}\angle AEF = \angle CEB \\AE = CE \\\angle EAF = \angle ECB\end{cases}$

 $\therefore \triangle AEF\cong\triangle CEB$（ASA）