連結 $AC$ 及 $AD$。
考慮 $\Delta ABC$。
$\begin{array}{rcll}
\angle ACB & = & \angle BAT & \text{(內錯弓形的圓周角)} \\
\angle ACB & = & 24^\circ
\end{array}$
另外，
$\begin{array}{rcll}
\angle CAE & = & \angle ABC & \text{(內錯弓形的圓周角)}
\end{array}$
由於 $AB = CD \text{(已知)}$，
$\begin{array}{rcll}
\therefore \overparen{AB} & = & \overparen{CD} & \text{(等弦對等弧)} \\
\therefore \angle CAD & = & \angle ACB & \text{(弧長與圓周角成比例)} \\
\angle CAD & = & 24^\circ
\end{array}$
考慮 $\Delta ADE$。
$\begin{array}{rcll}
\angle ADE & = & \angle ABC & \text{(圓內接四邊形的外角)} \\
\angle DAE & = & \angle CAE – \angle CAD \\
\angle DAE & = & \angle ABC – 24^\circ
\end{array}$
由此，可得
$\begin{array}{rcll}
\angle AED + \angle ADE + \angle DAE & = & 180^\circ & \text{($\Delta$ 的內角和)} \\
72^\circ +\angle ABC + \angle ABC – 24^\circ & = & 180^\circ \\
\angle ABC & = & 66^\circ
\end{array}$
