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第132页

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110°

4:3

 解：(1) 在$\triangle ABC$和$\triangle CED$中，

 $\begin{cases}AB = CE \\\angle B = \angle E \\BC = ED\end{cases}$

 $\therefore \triangle ABC\cong\triangle CED(SAS)，$$\therefore \angle CAB=\angle DCE，$$\therefore AB// CD。$

 (2) $\because \triangle ABC\cong\triangle CED，$$\therefore AB = CE = 5，$$AC = CD。$

 $\because AE = 2，$$\therefore CD = AC = CE - AE = 5 - 2 = 3。$

证明：(1) $\because BE$平分$\angle ABC，$$\therefore \angle FBE=\angle CBE。$

 $\because CE\perp BE，$$\therefore \angle FEB=\angle CEB = 90^{\circ}。$

 在$\triangle FBE$和$\triangle CBE$中，

 $\begin{cases}\angle FBE=\angle CBE \\BE = BE \\\angle FEB=\angle CEB\end{cases}$

 $\therefore \triangle FBE\cong\triangle CBE(ASA)，$$\therefore BF = BC。$

 (2) $\because \angle BAC=\angle BEF = 90^{\circ}，$$\therefore \angle CAF = 90^{\circ}，$$\angle ABD+\angle F=\angle ACF+\angle F = 90^{\circ}，$$\therefore \angle ABD=\angle ACF。$

 在$\triangle BDA$和$\triangle CFA$中，

 $\begin{cases}\angle ABD=\angle ACF \\AB = AC \\\angle BAD=\angle CAF = 90^{\circ}\end{cases}$

 $\therefore \triangle BDA\cong\triangle CFA(ASA)，$$\therefore BD = CF。$

 又$\because \triangle FBE\cong\triangle CBE，$$\therefore EF = EC，$即$CF = 2CE，$$\therefore BD = 2CE。$