In the given figure, AB ∥ DE and AC ∥ DF. Show that ABC ∼ DEF. If BC =10 cm, EB = CF =5 cm and AB =7 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 \(\displaystyle \triangle \mathrm{ABC} \sim \triangle \mathrm{DEF}\) \[\begin{aligned} & \frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{AB}}{\mathrm{DE}} \Rightarrow \frac{10}{20}=\frac{7}{\mathrm{DE}} \\ & \mathrm{DE}=14 \mathrm{~cm} \end{aligned} \]