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第23页

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95°

 （1）证明：

 在$\triangle ABE$和$\triangle CBD$中，

 $\begin{cases}AB = CB,\\\angle ABE = \angle CBD = 90^{\circ},\\BE = BD,\end{cases}$

 $\therefore \triangle ABE\cong\triangle CBD$（SAS）

 （2）解：

 $\because AB = CB，$$\angle ABC = 90^{\circ}，$

 $\therefore \triangle ABC$是等腰直角三角形.

 $\therefore \angle BAC = \angle ACB = 45^{\circ}.$

 $\because \triangle ABE\cong\triangle CBD，$

 $\therefore \angle AEB = \angle CDB.$

 $\because \angle AEB$为$\triangle AEC$的外角，

 $\therefore \angle AEB = \angle ACB + \angle CAE = 45^{\circ} + 30^{\circ} = 75^{\circ}.$

 $\therefore \angle CDB = 75^{\circ}$

 （1）证明：

 $\because \angle AOB = \angle COD = 90^{\circ}，$

 $\therefore \angle COD + \angle AOC = \angle AOB + \angle AOC，$即$\angle AOD = \angle BOC.$

 在$\triangle AOD$和$\triangle BOC$中，

 $\begin{cases}OA = OB,\\\angle AOD = \angle BOC,\\OD = OC,\end{cases}$

 $\therefore \triangle AOD\cong\triangle BOC$（SAS）

 （2）解：

 由（1），知$\triangle AOD\cong\triangle BOC，$

 $\therefore \angle A = \angle B.$

 设$OA，$$BC$交于点$F.$

 $\because \angle AOB = 90^{\circ}，$

 $\therefore \angle B + \angle OFB = 90^{\circ}.$

 又$\because \angle OFB = \angle AFE，$

 $\therefore \angle A + \angle AFE = 90^{\circ}.$

 $\therefore \angle AEB = 90^{\circ}$

 证明：

 $\because \angle ACB = 90^{\circ}，$$CF\perp AE，$

 $\therefore \angle ACF + \angle CAE = \angle ACF + \angle BCD = 90^{\circ}.$

 $\therefore \angle CAE = \angle BCD.$

 在$\triangle ACE$和$\triangle CBD$中，

 $\begin{cases}AC = CB,\\\angle CAE = \angle BCD,\\AE = CD,\end{cases}$

 $\therefore \triangle ACE\cong\triangle CBD$（SAS）.

 $\therefore \angle CBD = \angle ACE = 90^{\circ}，$即$BD\perp BC$