解：
(1) $\because AB=AC，$
$\therefore ∠ B = ∠ C。$
$\because ∠ ADC$是$△ ABD$的外角，
$\therefore ∠ ADC = ∠ B + ∠ BAD，$即$∠ ADE + ∠ CDE = ∠ B + ∠ BAD。$
$\because ∠ B = 40°，$$∠ ADE = 40°，$
$\therefore ∠ CDE = ∠ BAD。$
$\because AB=2，$$DC=2，$
$\therefore AB=DC。$
在$△ ABD$和$△ DCE$中，
$\begin{cases} ∠ BAD = ∠ CDE, \\ AB = DC, \\ ∠ B = ∠ C, \end{cases}$
$\therefore △ ABD ≌ △ DCE \ (\mathrm{ASA})。$
(2) 当$∠ BAD$的度数为$30°$或$60°$时，$△ ADE$是等腰三角形，理由如下：
在$△ ABC$中，$AB=AC，$
$\therefore ∠ B = ∠ C = 40°。$
① 若$AD=AE，$则$∠ AED = ∠ ADE = 40°。$
$\because ∠ AED$是$△ DEC$的外角，
$\therefore ∠ AED = ∠ EDC + ∠ C，$
$\therefore ∠ EDC = 0°，$此时点D、B重合，不符合题意，舍去。
② 若$AD=ED，$则$∠ DAE = ∠ DEA = \frac{1}{2}(180° - ∠ ADE) = \frac{1}{2} × (180° - 40°) =70°。$
$\because ∠ AED = ∠ EDC + ∠ C，$
$\therefore ∠ EDC = 30°，$
$\therefore ∠ BAD = ∠ EDC = 30°。$
③ 若$AE=DE，$则$∠ DAE = ∠ ADE = 40°。$
$\because △ ABC$的内角和为$180°，$
$\therefore ∠ BAC = 180° - 2×40° = 100°，$
$\therefore ∠ BAD = 100° - 40° = 60°。$
综上所述，当$∠ BAD$的度数为$30°$或$60°$时，$△ ADE$是等腰三角形。
