解：
(1) $\because$ 将$AD$绕点$A$按顺时针方向旋转$50°$能与线段$AE$重合，
$\therefore AD = AE，$$∠ DAE = 50°。$
$\because ∠ BAC = 50°，$
$\therefore ∠ BAC = ∠ DAE。$
$\therefore ∠ BAC - ∠ BAD = ∠ DAE - ∠ BAD，$即$∠ CAD = ∠ BAE。$
在$△ ACD$和$△ ABE$中，
$\begin{cases} AC = AB, \\ ∠ CAD = ∠ BAE, \\ AD = AE, \end{cases}$
$\therefore △ ACD ≌ △ ABE。$
$\therefore CD = BE。$
(2) $\because △ ACD ≌ △ ABE，$
$\therefore ∠ ADC = ∠ AEB。$
$\because ∠ ADC = 115°，$
$\therefore ∠ AEB = 115°。$
$\because AD = AE，$$∠ DAE = 50°，$
$\therefore ∠ AED = \frac{1}{2} × (180° - 50°) = 65°。$
$\therefore ∠ BED = ∠ AEB - ∠ AED = 50°。$
