考慮 $\Delta BDE$。
$\begin{array}{rcll}
BD & = & DE & \text{(已知)} \\
\angle DBE & = & \angle DEB & \text{(等腰 $\Delta$ 的底角)} \\
\end{array}$
所以，可得
$\begin{array}{rcll}
\angle DBE + \angle DEB & = & \angle ADB & \text{($\Delta$ 的外角)} \\
2\angle DEB & = & \angle ADB
\end{array}$
考慮四邊形 $ABCD$。
$\begin{array}{rcll}
\angle DAC & = &\angle DBE & \text{(同弓形內的圓周角)} \\
\angle DAC & = & \angle DEB & \text{(已證)}
\end{array}$
考慮 $\Delta ABD$。
$\begin{array}{rcll}
\angle ABD + \angle BAD + \angle ADB & = & 180^\circ & \text{($\Delta$ 的內角和)} \\
\angle ABD + \angle BAC + \angle CAD + \angle ADB & = & 180^\circ \\
30^\circ + 66^\circ + \angle DEB + 2 \angle DEB & = & 180^\circ & \text{(已證)} \\
3 \angle DEB & = & 84^\circ \\
\angle DEB & = & 28^\circ \\
\angle CED & = & 28^\circ
\end{array}$
