In the given figure, $DE \| AC$ and $DF \| AE$.
Prove that $ \frac{\mathbf{B F}}{\mathbf{F E}}=\frac{\mathbf{B E}}{\mathbf{E C}} $
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Given:

$DE \| AC$ and $DF \| AE$.

To do:

We have to prove that \( \frac{\mathbf{B F}}{\mathbf{F E}}=\frac{\mathbf{B E}}{\mathbf{E C}} \)

Solution:

We know that,

If a line divides two sides of a triangle proportionally, then it is parallel to the third side.

In $\triangle ABC, DE \| AC$,

This implies,

$\frac{BD}{AD}=\frac{BE}{EC}$.........(i)

In $\triangle ABE, DF \| AE$,

This implies,

$\frac{BD}{AD}=\frac{BF}{EF}$.........(ii)

From (i) and (ii), we get,

$\frac{BF}{FE}=\frac{BE}{EC}$

Hence proved.