jmc

Since $AB\parallel EF,$ we know that $\angle BAC=\angle FEC$ and $\angle ABC=\angle EFC.$ Therefore, we see that $△ABC\sim △EFC$ by AA Similarity. Likewise, $△BDC\sim △BEF.$

From our similarities, we can come up with two equations: $\frac{BF}{BC}=\frac{EF}{DC}$ and $\frac{FC}{BC}=\frac{EF}{AB}.$

Since we have $AB$ and $DC$ and we want to find $EF,$ we want all the other quantities to disappear. Since $BF+FC=BC,$ we try adding our two equations: $$\begin{array}{rl}\frac{BF}{BC}+\frac{FC}{BC}& =\frac{EF}{DC}+\frac{EF}{AB}.\\ \frac{BC}{BC}=1& =EF\left(\frac{1}{DC}+\frac{1}{AB}\right)\\ \frac{1}{\frac{1}{DC}+\frac{1}{AB}}& =EF\end{array}$$ Now we plug in $DC=100\text{ cm}$ and $AB=150\text{ cm},$ giving us $EF=\boxed{60}\text{ cm}.$