Method 1:
In $\Delta ACD$ and $\Delta BDE$,
$\angle CAD = \angle DBE = 60^\circ$ (properties of equil. $\Delta$)
$\begin{array}{rcll}
\angle ACD & = & \angle CDB – \angle CAD & \text{(ext. $\angle$ of $\Delta$)} \\
\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)
Therefore, we have
$\begin{array}{rcll}
\dfrac{AC}{BD} & = & \dfrac{AD}{BE} & \text{(corr. sides, $\sim \Delta$)} \\
\dfrac{16}{16-4} & = & \dfrac{4}{BE} & \\
BE & = & 3
\end{array}$
Hence, we have
$\begin{array}{rcl}
CE & = & BC – BE \\
CE & = & 16 – 3 \\
CE & = & 13 \text{ cm}
\end{array}$
Method 2:
By applying cosine law to $\Delta ACD$, we have
$\begin{array}{rcl}
CD^2 & = & AC^2 + AD^2 – 2 (AC) (AD) \cos \angle CAD \\
CD^2 & = & (16)^2 + (4)^2 – 2(16)(4) \cos 60^\circ \\
CD^2 & = & 208 \\
CD & = & \sqrt{208} \text{ cm}
\end{array}$
By applying sine law to $\Delta BCD$, we have
$\begin{array}{rcl}
\dfrac{BD}{\sin \angle BCD} & = & \dfrac{CD}{\sin \angle CBD} \\
\dfrac{12}{\sin \angle BCD} & = & \dfrac{\sqrt{208}}{\sin 60^\circ} \\
\angle BCD & = & 46.102\ 113\ 75^\circ
\end{array}$
Consider $\Delta CDE$, we have
$\begin{array}{cl}
& \angle CED \\
= & 180^\circ – \angle CDE – \angle DCE \\
= & 180^\circ – 60^\circ – 46.102\ 113\ 75^\circ \\
= & 73. 897\ 886\ 25^\circ
\end{array}$
By applying sine law to $\Delta CDE$, we have
$\begin{array}{rcl}
\dfrac{CE}{\sin \angle CDE} & = & \dfrac{CD}{\sin \angle CED} \\
\dfrac{CE}{\sin 60^\circ} & = & \dfrac{\sqrt{208}}{\sin 73. 897\ 886\ 25^\circ} \\
CE & = & 13 \text{ cm}
\end{array}$
