（1）解：在$\triangle ABC$和$\triangle ADE$中，
$\begin{cases}AB = AD \\BC = DE \\AC = AE\end{cases}$
$\therefore \triangle ABC\cong\triangle ADE(SSS)$
$\therefore \angle BAC=\angle DAE$
$\therefore \angle BAC - \angle DAC=\angle DAE - \angle DAC，$即$\angle BAD=\angle CAE = 38^{\circ}$
（2）证明：由（1）知，$\triangle ABC\cong\triangle ADE，$$\angle BAD=\angle CAE$
$\therefore \angle C=\angle E$
又$\because \angle E+\angle AFE+\angle CAE = 180^{\circ}，$$\angle C+\angle CFD+\angle CDE = 180^{\circ}，$$\angle AFE=\angle CFD$
$\therefore \angle CDE=\angle CAE$
$\because \angle BAD=\angle CAE$
$\therefore \angle CDE=\angle BAD$
