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$40^{\circ}$

解：（1）如图，BC的垂直平分线l，点E，F，G即为所求 

（2）如图，设BE与AM交于点D. $\because$ EF是BC的垂直平分线，$\therefore$ EB = EC，$\angle GFM = 90^{\circ}.$ $\therefore$ $\angle EBC = \angle C.$ $\because$ $\angle AGE = \angle C，$$\therefore$ $\angle EBC = \angle AGE.$ $\because$ $\angle GMF = \angle BMD，$$\therefore$ $\angle BDM = \angle GFM = 90^{\circ}.$ 

$\therefore$ AG $\perp$ BE. $\therefore$ $\angle BDA = \angle EDA = 90^{\circ}.$

$\because$ AM平分$\angle BAC，$$\therefore$ $\angle BAD = \angle EAD.$

$\because$ AD = AD，$\therefore$ $\triangle ABD \cong \triangle AED.$

$\therefore$ BD = ED. $\therefore$ AG垂直平分BE

 解：（1）$\because$ AD$//$BE，$\therefore$ $\angle ADB = \angle DBC.$ 

$\because$ BD平分$\angle ABC，$$\therefore$ $\angle ABD = \angle DBC.$ 

$\therefore$ $\angle ABD = \angle ADB.$ $\therefore$ AB = AD 

（2）$\because$ AD$//$BE，$\therefore$ $\angle ADC = \angle DCE.$ 

由（1），知AB = AD，

又$\because$ AB = AC，$\therefore$ AC = AD.

 $\therefore$ $\angle ACD = \angle ADC.$ $\therefore$ $\angle ACD = \angle DCE.$ 

$\therefore$ CD平分$\angle ACE$