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第64页

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$ -2$

$ 4$

$ 2$

 解：（1）连接AE. 

$\because$ EF垂直平分AB，$\therefore$ AE = BE. 

$\because$ BE = AC，$\therefore$ AE = AC. 

$\because$ D是EC的中点，$\therefore$ AD $\perp$ BC 

（2）设$\angle B = x^{\circ}.$ $\because$ AE = BE，

$\therefore$ $\angle BAE = \angle B = x^{\circ}.$ $\therefore$ $\angle AEC = 2x^{\circ}.$ 

$\because$ AE = AC，$\therefore$ $\angle C = \angle AEC = 2x^{\circ}.$ $\triangle ABC$$x^{\circ}+2x^{\circ}+75^{\circ}=180^{\circ}，$
$\therefore$$x = 35.$$\therefore$

在$\triangle ABC$中，$x^{\circ}+2x^{\circ}+75^{\circ}=180^{\circ}，$
$\therefore$ $x = 35.$ $\therefore$ $\angle B = 35^{\circ}$

 解：（1）如图①，连接AD. 

$\because$ AB = AC，$\angle BAC = 90^{\circ}，$D为BC的中点，

$\therefore$ AD $\perp$ BC，$\angle B = \angle C = 45^{\circ}，$$\angle BAD = \angle DAC = 45^{\circ}.$ $\therefore$ $\angle BDA = 90^{\circ}，$$\angle B = \angle BAD = \angle DAC.$ 

$\therefore$ BD = AD. 又$\because$ BE = AF，

$\therefore$ $\triangle BDE \cong \triangle ADF.$ 

$\therefore$ ED = FD，$\angle BDE = \angle ADF.$ 

$\therefore$ $\angle EDF = \angle EDA + \angle ADF = \angle EDA + \angle BDE = \angle BDA = 90^{\circ}.$ 

$\therefore$ $\triangle DEF$为等腰直角三角形 

（2）$\triangle DEF$仍为等腰直角三角形 理由：如图②，

连接AD. 

$\because$ AB = AC，$\angle BAC = 90^{\circ}，$D为BC的中点，

$\therefore$ AD $\perp$ BC，$\angle DAC = \angle BAD = \angle ABD = 45^{\circ}.$ 

$\therefore$ $\angle BDA = 90^{\circ}，$AD = BD，$\angle DAF = \angle DBE = 135^{\circ}.$ 

又$\because$ AF = BE，$\therefore$ $\triangle DAF \cong \triangle DBE.$ 

$\therefore$ FD = ED，$\angle ADF = \angle BDE.$ 

$\therefore$ $\angle EDF = \angle BDE + \angle FDB = \angle ADF+\angle FDB = \angle BDA = 90^{\circ}.$ 

$\therefore$ $\triangle DEF$仍为等腰直角三角形