State the converse of basic proportionality theorem. Also find $\displaystyle \frac{\mathrm{BF}}{\mathrm{FC}}$ in the following figure, given that $\displaystyle \mathrm{AB}\|\mathrm{DC}\| \mathrm{EF}$ and $\displaystyle \frac{\mathrm{AE}}{\mathrm{ED}}=\frac{2}{3}$. Also, find the length of EF if $\displaystyle \mathrm{AB}=10 \mathrm{~cm}$ and $\displaystyle \mathrm{DC}=15 \mathrm{~cm}$.

Correct statement of converse of Basic Proportionality Theorem. In \(\displaystyle \Delta \mathrm{ADC}, \mathrm{EG} \| \mathrm{DC}\) \[\Rightarrow \frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{AG}}{\mathrm{GC}}=\frac{2}{3} \] In \(\displaystyle \Delta \mathrm{ABC}, \mathrm{GF} \| \mathrm{AB}\) \[\Rightarrow \frac{\mathrm{AG}}{\mathrm{GC}}=\frac{\mathrm{BF}}{\mathrm{FC}}=\frac{2}{2} \] \(\displaystyle \Delta \mathrm{AEG} \sim \Delta \mathrm{ADC}\) \[\begin{aligned} & \Rightarrow \frac{\mathrm{AE}}{\mathrm{AD}}=\frac{\mathrm{AG}}{\mathrm{AC}}=\frac{\mathrm{EG}}{\mathrm{DC}} \\ & \Rightarrow \frac{2}{5}=\frac{\mathrm{EG}}{\mathrm{DC}} \\ & \Rightarrow \mathrm{EG}=\frac{2}{5} \times 15=6 \mathrm{~cm} \\ & \text { Similarly, } \Delta \mathrm{CFG} \sim \Delta \mathrm{CBA} \text { and } \frac{\mathrm{FC}}{\mathrm{BF}}=\frac{3}{2} \\ & \Rightarrow \frac{\mathrm{FC}}{\mathrm{BC}}=\frac{\mathrm{GF}}{\mathrm{AB}}=\frac{3}{5} \\ & \Rightarrow \mathrm{GF}=\frac{3}{5} \times 10=6 \mathrm{~cm} \\ & \mathrm{EF}=\mathrm{EG}+\mathrm{GF}=6+6=12 \mathrm{~cm} \end{aligned} \] \[\begin{aligned} & \Rightarrow \frac{\mathrm{FC}}{\mathrm{BC}}=\frac{\mathrm{GF}}{\mathrm{AB}}=\frac{3}{5} \\ & \Rightarrow \mathrm{GF}=\frac{3}{5} \times 10=6 \mathrm{~cm} \end{aligned} \]