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

Question.

In figure, $D E \| A C$ and $D F \| A E$. Prove that $\frac{B F}{F E}=\frac{B E}{E C}$.

Solution:

In $\triangle \mathrm{ABE}$

$\mathrm{DF} \| \mathrm{AE}$ (Given)

$\frac{B D}{D A}=\frac{B F}{F E} \ldots$ (i) (Basic Proportionality Theorem)

In $\triangle \mathrm{ABC}$,

$D E \| A C$(Given)

$\frac{B D}{D A}=\frac{B E}{E C}$ (ii) (Basic Proportionality Theorem)

From (i) and (ii), we get

$\frac{B F}{F E}=\frac{B E}{E C} \quad$ Hence proved.