Question

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

Hint:

When multiple parallel lines are given within nested triangles, look for a common ratio (like \(BD/DA\) here) that links the different parts of the segments together.

Answer 1

Step 1: Apply BPT in Triangle ABE
Given DF ∥ AE

In ΔABE, a line parallel to AE cuts sides AB and BE at D and F.

By Basic Proportionality Theorem:
BD / DA = BF / FE …(1)

Step 2: Apply BPT in Triangle ABC
Given DE ∥ AC

In ΔABC, a line parallel to AC cuts sides AB and BC at D and E.

By Basic Proportionality Theorem:
BD / DA = BE / EC …(2)

Step 3: Compare Equations
From (1) and (2):
BD / DA is common

So,
BF / FE = BE / EC

Final Answer:
Hence proved that BF / FE = BE / EC.