jmc

Since $DE\parallel AB,$ we know that $\angle CDE=\angle CAB$ and $\angle CED=\angle CBA.$ Therefore, by AA similarity, we have $△ABC\sim DEC.$ Then, we find: $$\begin{array}{rl}\frac{CB}{CE}& =\frac{CA}{CD}=\frac{CD+DA}{CD}\\ \frac{CB}{6\text{ cm}}& =\frac{4\text{ cm}+10\text{ cm}}{4\text{ cm}}=\frac{7}{2}\\ CB& =6\text{cm}\cdot \frac{7}{2}=\boxed{21}\text{ cm}.\end{array}$$