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

We know that,

If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, the other two sides are divided in the same ratio.

In Δ ABC,

DE || AC [∵ Given]

$\therefore \dfrac{BD}{AD} = \dfrac{BE}{EC}$ ………(1)

In Δ ABE,

DF || AE [∵ Given]

$\therefore \dfrac{BD}{AD} = \dfrac{BF}{FE}$ ……..(2)

From (1) and (2), we get :

$\Rightarrow \dfrac{BE}{EC} = \dfrac{BF}{FE}$

Hence, proved that $\dfrac{BF}{FE} = \dfrac{BE}{EC}$.