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

(N/A) In $\Delta ABC$,$DE \parallel AC$.
By the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{BD}{DA} = \frac{BE}{EC} \quad ...(i)$
In $\Delta BAE$,$DF \parallel AE$.
By the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{BD}{DA} = \frac{BF}{FE} \quad ...(ii)$
From equations $(i)$ and $(ii)$,since the left-hand sides are equal,the right-hand sides must also be equal:
$\frac{BF}{FE} = \frac{BE}{EC}$
Hence,it is proved.