In the given Figure 5, $DE \parallel AC$ and $DF \parallel AE$. Prove that $\frac{BF}{FE} = \frac{BE}{EC}$.

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Question

In the given Figure 5, $DE \parallel AC$ and $DF \parallel AE$. Prove that $\frac{BF}{FE} = \frac{BE}{EC}$.

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Accepted Answer

Step 1: In $	riangle ABE$, $DF \parallel AE$, so by BPT: $\frac{BD}{DA} = \frac{BF}{FE}$ ... (i)
Step 2: In $	riangle ABC$, $DE \parallel AC$, so by BPT: $\frac{BD}{DA} = \frac{BE}{EC}$ ... (ii)
Step 3: From (i) and (ii): $\frac{BF}{FE} = \frac{BE}{EC}$

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