In $\Delta CDE$,
$\begin{array}{rcll}
\angle CED & = & \angle AEB & \text{(vert. opp. $\angle$s)} \\
\angle CED & = & 94^\circ & \\
\end{array}$
Also,
$\begin{array}{rcll}
\angle DCE & = & 180^\circ -\angle CDE -\angle CED & \text{($\angle$ sum of $\Delta$)} \\
\angle DCE & = & 180^\circ -28^\circ -94^\circ & \\
\angle DCE & = & 58^\circ
\end{array}$
Since $AB\text{//}CD$, we have
$\begin{array}{rcll}
\angle ABC & = & \angle BCD & \text{(alt. $\angle$s, $AB$//$CD$)} \\
\angle ABC & = & 58^\circ
\end{array}$
Also,
$\begin{array}{rcll}
\angle BAD & = & \angle ADC & \text{(alt. $\angle$s, $AB$//$CD$)} \\
\angle BAD & = & 28^\circ
\end{array}$
In $\Delta ABC$,
$\begin{array}{rcll}
\because AB & = & BC & \text{(given)} \\
\therefore \angle BAC & = & \angle BCA & \text{(base $\angle$s, isos. $\Delta$)} \\
\angle BAC & = & \dfrac{1}{2}(180^\circ -\angle ABC) & \text{($\angle$ sum of $\Delta$)} \\
\angle BAC & = & \dfrac{1}{2}(180^\circ -58^\circ) & \\
\angle BAC & = & 61^\circ
\end{array}$
Hence, we have
$\begin{array}{rcl}
\angle CAD & = & \angle BAC – \angle BAD \\
\angle CAD & = & 61^\circ -28^\circ \\
\angle CAD & = & 33^\circ
\end{array}$
