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 解：（1）证明：

在$\triangle ABC$和$\triangle ADE$中，

$\begin{cases}BC = DE\\\angle B = \angle D\\AB = AD\end{cases}，$
$\therefore\triangle ABC\cong\triangle ADE。$

（2）由（1），得$\triangle ABC\cong\triangle ADE，$
$\therefore AC = AE，$$\angle BAC = \angle DAE = 60^{\circ}。$
$\therefore\angle AEC = \angle ACE。$
$\because\angle AEC+\angle ACE = 2\angle ACE = 180^{\circ}-\angle DAE = 120^{\circ}，$
$\therefore\angle ACE = 60^{\circ}。$

 解：（1）证明：$\because\angle A=\angle A，$$\angle ABE=\angle ACF，$$AE = AF，$
$\therefore\triangle ABE\cong\triangle ACF。$$\therefore AB = AC。$
$\therefore\angle ABC = \angle ACB。$
$\therefore\angle ABC-\angle ABE=\angle ACB-\angle ACF，$即$\angle DBC = \angle DCB。$
$\therefore BD = CD，$即$\triangle BCD$是等腰三角形。

（2）由（1），知$AB = AC，$

又$\because\angle A = 40^{\circ}，$
$\therefore\angle ABC=\angle ACB=\dfrac{1}{2}\times(180^{\circ}-40^{\circ}) = 70^{\circ}。$
$\because BD = CD，$$BC = BD，$
$\therefore BD = CD = BC。$
$\therefore\triangle DBC$是等边三角形。

$\therefore\angle DBC = 60^{\circ}。$
$\therefore\angle ABE=\angle ABC-\angle DBC = 10^{\circ}。$
$\therefore\angle BEC=\angle A+\angle ABE = 50^{\circ}。$