$\begin{array}{rcll}
\angle ACD & = & 180^\circ -\angle AED & \text{(opp. $\angle$s, cyclic quad.)} \\
\angle ACD & = & 180^\circ -96^\circ \\
\angle ACD & = & 84^\circ
\end{array}$
Join $AB$.
$\begin{array}{rcll}
AC & = & BD & \text{(given)} \\
\overparen{AC} & = & \overparen{BD} & \text{(eq. chords, eq. arcs)} \\
\angle ADC & = & \angle BAD & \text{(arcs prop. to $\angle$s at $\unicode{x2299}^{ce}$)} \\
\because \angle CDB & = & \angle BAC & \text{($\angle$s in the same segment)} \\
\therefore \angle ADB & = & \angle CAD
\end{array}$
In $\Delta ACD$,
$\begin{array}{rcll}
\angle CAD +\angle ADB +\angle BDC +\angle ACD & = & 180^\circ & \text{($\angle$ sum of $\Delta$)} \\
\angle CAD +\angle CAD +14^\circ +84^\circ & = & 180^\circ \\
\angle CAD & = & 41^\circ
\end{array}$
