答案：A
$\begin{array}{rcll}
\angle ACD & = & 180^\circ -\angle AED & \text{(圓內接四邊形的對角)} \\
\angle ACD & = & 180^\circ -96^\circ \\
\angle ACD & = & 84^\circ
\end{array}$
連結 $AB$。
$\begin{array}{rcll}
AC & = & BD & \text{(given)} \\
\overparen{AC} & = & \overparen{BD} & \text{(等弦對等弧)} \\
\angle ADC & = & \angle BAD & \text{(弧長與圓周角成比例)} \\
\because \angle CDB & = & \angle BAC & \text{(同弓形內的圓周角)} \\
\therefore \angle ADB & = & \angle CAD
\end{array}$
在 $\Delta ACD$ 中，
$\begin{array}{rcll}
\angle CAD +\angle ADB +\angle BDC +\angle ACD & = & 180^\circ & \text{(三角形的內角和)} \\
\angle CAD +\angle CAD +14^\circ +84^\circ & = & 180^\circ \\
\angle CAD & = & 41^\circ
\end{array}$
