23. Answer: 8 .

We have
$$
\begin{array}{l}
\angle \mathrm{ABE}=180^{\circ}-60^{\circ}=120^{\circ}=\angle \mathrm{ECD} \text { and } \\
\angle \mathrm{AEB}=180^{\circ}-\left(\angle \mathrm{BAE}+120^{\circ}\right)=\angle \mathrm{CDE} .
\end{array}
$$

Therefore $\triangle \mathrm{ABE}$ is similar to $\triangle \mathrm{ECD}$, and it follows that $\mathrm{AB}: \mathrm{BE}=\mathrm{EC}: C D$. Hence $B C^{2}=(B E)(E C)=(A B)(C D)=64 \mathrm{~cm}^{2}$, so $B C=8 \mathrm{~cm}$.
23. In the figure below, $\mathrm{ADE}$ is a triangle with $\angle \mathrm{AED}=120^{\circ}$, and $\mathrm{B}$ and $\mathrm{C}$ are points on side $\mathrm{AD}$ such that $\mathrm{BCE}$ is an equilateral triangle. If $\mathrm{AB}=4 \mathrm{~cm}, \mathrm{CD}=16 \mathrm{~cm}$ and $\mathrm{BC}=x \mathrm{~cm}$, find the value of $x$.