In the given figure, $\displaystyle \mathrm{DE} \| \mathrm{AC}$ and $\displaystyle \mathrm{DF} \| \mathrm{AE}$. Prove that : $\displaystyle \frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}$.

In \(\displaystyle \triangle \mathrm{BEA}, \mathrm{FD} \| \mathrm{EA}\) \[\therefore \frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BD}}{\mathrm{DA}} \] In \(\displaystyle \triangle \mathrm{BCA}, \mathrm{ED} \| \mathrm{CA}\) \[\therefore \frac{\mathrm{BE}}{\mathrm{EC}}=\frac{\mathrm{BD}}{\mathrm{DA}} \] Using (i) and (ii) \[\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{E C} \]