In the given figure, AB $\displaystyle \| \mathrm{DE}$ and AC $\displaystyle \| \mathrm{DF}$. Show that $\displaystyle \triangle \mathrm{ABC} \sim \triangle \mathrm{DEF}$. If $\displaystyle \mathrm{BC}=10 \mathrm{~cm}, \mathrm{~EB}=\mathrm{CF}=5 \mathrm{~cm}$ and $\displaystyle \mathrm{AB}=7 \mathrm{~cm}$, then find the length DE.

\(\displaystyle \mathrm{AB} \| \mathrm{DE} \Rightarrow \angle \mathrm{DEF}=\angle \mathrm{ABC}\) \(\displaystyle \mathrm{AC} \| \mathrm{DF} \Rightarrow \angle \mathrm{DFE}=\angle \mathrm{ACB}\) Hence \(\displaystyle \triangle \mathrm{ABC} \sim \triangle \mathrm{DEF}\) \[\begin{aligned} & \frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{AB}}{\mathrm{DE}} \Rightarrow \frac{10}{20}=\frac{7}{\mathrm{DE}} \\ & \mathrm{DE}=14 \mathrm{~cm} \end{aligned} \]