Consider $\Delta BDE$.
$\begin{array}{rcll}
BD & = & DE & \text{(given)} \\
\angle DBE & = & \angle DEB & \text{(base $\angle$s, isos. $\Delta$)} \\
\end{array}$
Then, we have
$\begin{array}{rcll}
\angle DBE + \angle DEB & = & \angle ADB & \text{(ext. $\angle$ of $\Delta$)} \\
2\angle DEB & = & \angle ADB
\end{array}$
Consider quadrilateral $ABCD$.
$\begin{array}{rcll}
\angle DAC & = &\angle DBE & \text{($\angle$s in the same segment)} \\
\angle DAC & = & \angle DEB & \text{(proved)}
\end{array}$
Consider $\Delta ABD$.
$\begin{array}{rcll}
\angle ABD + \angle BAD + \angle ADB & = & 180^\circ & \text{($\angle$ sum of $\Delta$)} \\
\angle ABD + \angle BAC + \angle CAD + \angle ADB & = & 180^\circ \\
30^\circ + 66^\circ + \angle DEB + 2 \angle DEB & = & 180^\circ & \text{(proved)} \\
3 \angle DEB & = & 84^\circ \\
\angle DEB & = & 28^\circ \\
\angle CED & = & 28^\circ
\end{array}$
