解：（1）如图所示。
（2）$\because\angle BAC = 40^{\circ}，$$AB = AC，$
$\therefore\angle ABC=\angle ACB=\dfrac{1}{2}\times(180^{\circ}-40^{\circ}) = 70^{\circ}。$
由（1），知$AD\perp BF，$$AB = AF，$
$\therefore BD = FD。$$\therefore\angle ABD=\angle AFD = 70^{\circ}。$
$\therefore\angle BDF = 40^{\circ}。$
（3）如图，连接$FC。$
$\because AD\perp BF，$$AB = AF，$
$\therefore DB = DF。$
$\because DC = DE，$
$\therefore\angle DCE=\angle DEC=\angle ACB=\angle ABC=\angle DFB。$
$\therefore\angle BDF=\angle BDE。$
$\therefore\triangle DCF\cong\triangle DEB。$
$\therefore\angle DEB=\angle DCF。$
$\because AB = AC = AF，$
$\therefore$易得$\angle FCB=\angle FCD = 90^{\circ}。$
$\therefore\angle DEB = 90^{\circ}。$$\therefore BE\perp DE。$
