In $\Delta ABD$ and $\Delta CAE$,
$\begin{array}{rcll}
AB & = & CA & \text{(equilateral triangle)} \\
AD & = & CE & \text{(given)} \\
\angle BAD & = & \angle ACE & \text{(equilateral triangle)}
\end{array}$
$\therefore \Delta ABD \cong \Delta CAE$ (S.A.S.).
Hence, we have
$\begin{array}{rcll}
\angle EAC & = & \angle DBA & \text{(corr. $\angle$s, $\cong\Delta$s)} \\
\angle EAC & = & \angle ABC -\angle CBD \\
\angle EAC & = & 60^\circ -38^\circ & \text{(equilateral triangle)} \\
\angle EAC & = & 22^\circ
\end{array}$
In $\Delta ACE$,
$\begin{array}{rcll}
\angle AEB & = & \angle CAE +\angle ACE & \text{(ext. $\angle$ of $\Delta$)} \\
\angle AEB & = & 22^\circ +60^\circ & \text{(equilateral triangle)} \\
\angle AEB & = & 82^\circ
\end{array}$
