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: Understanding the Concept:
Basic Proportionality Theorem (Thales's Theorem) states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Step 2: Key Formula or Approach:
In \(\triangle XYZ\), if \(PQ \parallel YZ\), then \(\frac{XP}{PY} = \frac{XQ}{QZ}\).
Step 3: Detailed Explanation:
In \(\triangle ABE\), it is given that \(DF \parallel AE\).
Applying Basic Proportionality Theorem:
\[ \frac{BD}{DA} = \frac{BF}{FE} \quad \text{--- (i)} \]
In \(\triangle ABC\), it is given that \(DE \parallel AC\).
Applying Basic Proportionality Theorem:
\[ \frac{BD}{DA} = \frac{BE}{EC} \quad \text{--- (ii)} \]
From equations (i) and (ii), we observe that the Left Hand Side (LHS) is identical for both:
\[ \frac{BF}{FE} = \frac{BE}{EC} \]
Step 4: Final Answer:
Hence proved.