电子课本网 › 第51页

第51页

信息发布者：

$35^{\circ}$

$67^{\circ}$

 证明：$\because AB = AC，$$AD$是边$BC$上的中线，

$\therefore AD\perp BC，$$\angle BAD=\angle CAD。$
$\therefore \angle CAD+\angle C = 90^{\circ}。$$\because BE\perp AC，$
$\therefore \angle CBE+\angle C=90^{\circ}。$$\therefore \angle CBE=\angle CAD。$
$\therefore \angle CBE=\angle BAD$

 证明：连接$DE，$$DF。$

$\because AB = AC，$$\therefore \angle B=\angle C。$

在$\triangle BDE$和$\triangle CFD$中，

$\begin{cases}BE = CD\\\angle B=\angle C\\BD = CF\end{cases}，$
$\therefore \triangle BDE\cong\triangle CFD。$$\therefore DE = FD。$
$\because G$是$EF$的中点，$\therefore DG\perp EF$

 解：（1）$\because \angle BAC = 90^{\circ}，$$AB = AC，$
$\therefore \angle B=\angle C=\frac{1}{2}(180^{\circ}-\angle BAC)=45^{\circ}。$

又$\because \angle BAD = 30^{\circ}，$$\therefore \angle ADC=\angle B+\angle BAD=45^{\circ}+30^{\circ}=75^{\circ}。$
$\because \angle DAC=\angle BAC-\angle BAD=90^{\circ}-30^{\circ}=60^{\circ}，$且$AD = AE，$
$\therefore \angle ADE=\angle AED=\frac{1}{2}(180^{\circ}-\angle DAC)=60^{\circ}。$
$\therefore \angle EDC=\angle ADC-\angle ADE=75^{\circ}-60^{\circ}=15^{\circ}$ 

（2）$\because \angle BAC=\alpha，$$\angle BAD = 30^{\circ}，$$\therefore \angle DAC=\alpha - 30^{\circ}。$
$\because AB = AC，$$\therefore \angle B=\angle C=\frac{1}{2}(180^{\circ}-\angle BAC)=90^{\circ}-\frac{1}{2}\alpha。$
$\therefore \angle ADC=\angle B+\angle BAD=120^{\circ}-\frac{1}{2}\alpha。$
$\because AD = AE，$$\therefore \angle ADE=\angle AED=\frac{1}{2}(180^{\circ}-\angle DAC)=105^{\circ}-\frac{1}{2}\alpha。$
$\therefore \angle EDC=\angle ADC-\angle ADE=(120^{\circ}-\frac{1}{2}\alpha)-(105^{\circ}-\frac{1}{2}\alpha)=15^{\circ}$ 

（3）$\angle EDC=\frac{1}{2}\angle BAD$