方法 1:
在 $\Delta ACD$ 及 $\Delta BDE$ 中，
$\angle CAD = \angle DBE = 60^\circ$ (等邊 $\Delta$ 的性質)
$\begin{array}{rcll}
\angle ACD & = & \angle CDB – \angle CAD & \text{($\Delta$ 的外角)} \\
\angle ACD & = & \angle CDE + \angle BDE – \angle CAD & \\
\angle ACD & = & 60^\circ + \angle BDE – 60^\circ & \\
\angle ACD & = & \angle BDE
\end{array}$
$\therefore \Delta ACD \sim \Delta BDE$ (A.A.A)
所以，可得
$\begin{array}{rcll}
\dfrac{AC}{BD} & = & \dfrac{AD}{BE} & \text{($\sim \Delta$ 的對應邊)} \\
\dfrac{16}{16-4} & = & \dfrac{4}{BE} & \\
BE & = & 3
\end{array}$
由此，可得
$\begin{array}{rcl}
CE & = & BC – BE \\
CE & = & 16 – 3 \\
CE & = & 13 \text{ cm}
\end{array}$
方法 2:
在 $\Delta ACD$ 運用餘弦公式，可得
$\begin{array}{rcl}
CD^2 & = & AC^2 + AD^2 – 2 (AC) (AD) \cos \angle CAD \\
CD^2 & = & (16)^2 + (4)^2 – 2(16)(4) \cos 60^\circ \\
CD^2 & = & 208 \\
CD & = & \sqrt{208} \text{ cm}
\end{array}$
在 $\Delta BCD$ 中運用正弦公式，可得
$\begin{array}{rcl}
\dfrac{BD}{\sin \angle BCD} & = & \dfrac{CD}{\sin \angle CBD} \\
\dfrac{12}{\sin \angle BCD} & = & \dfrac{\sqrt{208}}{\sin 60^\circ} \\
\angle BCD & = & 46.102\ 113\ 75^\circ
\end{array}$
考慮 $\Delta CDE$，可得
$\begin{array}{cl}
& \angle CED \\
= & 180^\circ – \angle CDE – \angle DCE \\
= & 180^\circ – 60^\circ – 46.102\ 113\ 75^\circ \\
= & 73. 897\ 886\ 25^\circ
\end{array}$
在 $\Delta CDE$ 中運用正弦公式，可得
$\begin{array}{rcl}
\dfrac{CE}{\sin \angle CDE} & = & \dfrac{CD}{\sin \angle CED} \\
\dfrac{CE}{\sin 60^\circ} & = & \dfrac{\sqrt{208}}{\sin 73. 897\ 886\ 25^\circ} \\
CE & = & 13 \text{ cm}
\end{array}$
