Let $x = \angle ACD$.
In $\Delta BCD$,
$\begin{array}{rcll}
BC & = & CD & \text{(given)} \\
\angle CBD & = & \angle BDC & \text{(base $\angle$s, isos. $\Delta$)} \\
\angle CBD & = & \dfrac{1}{2} (180^\circ – x) & \text{($\angle$ sum of $\Delta$)}
\end{array}$
In $\Delta ABD$,
$\begin{array}{rcll}
AB & = & BD & \text{(given)} \\
\angle BAD & = & \angle BDA & \text{(base $\angle$s, isos. $\Delta$)} \\
2\angle BAD & = & \angle CBD & \text{(ext. $\angle$ of $\Delta$)} \\
2\angle BAD & = & \dfrac{1}{2} (180^\circ – x) \\
\angle BAD & = & \dfrac{1}{4} (180^\circ – x) \\
\end{array}$
In $\Delta ACD$,
$\begin{array}{rcll}
\angle CAD + \angle ACD & = & \angle CDE & \text{(ext. $\angle$ of $\Delta$)} \\
\dfrac{1}{4} ( 180^\circ – x) + x & = & 66^\circ \\
180^\circ – x + 4x & = & 264^\circ \\
3x & = & 84^\circ \\
x & = & 28^\circ
\end{array}$
