Question

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

Hint:

Whenever a geometry proof involves proving the equality of two ratios, look for a common third ratio that is shared by both parts of the diagram.
In this problem, the ratio of the segments on the side \(AB\) (\(\frac{BD}{DA}\)) serves as the bridge connecting both proportions!

Answer 1

Step 1: Understanding the Question:
The topic of this question is Triangles, specifically the Basic Proportionality Theorem (also known as Thales' Theorem).
We are given a triangle \(ABC\) with points \(D\) on \(AB\), and \(E, F\) on \(BC\).
We are given two parallel relationships: \(DE \parallel AC\) and \(DF \parallel AE\).
We need to prove the ratio equality \(\frac{BF}{FE} = \frac{BE}{EC}\).

Step 2: Key Formula or Approach:
By the Basic Proportionality Theorem (BPT), if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
We will apply BPT to two different triangles inside our main figure:
1. \(\Delta ABC\) with \(DE \parallel AC\)
2. \(\Delta ABE\) with \(DF \parallel AE\)
We will then equate the common ratio to complete our proof.

Step 3: Detailed Explanation:

• Consider the larger triangle \(\Delta ABC\):
We are given that:
\[ DE \parallel AC \] Therefore, by applying the Basic Proportionality Theorem to \(\Delta ABC\), we get:
\[ \frac{BD}{DA} = \frac{BE}{EC} \quad \text{--- (Equation 1)} \]

• Consider the smaller triangle \(\Delta ABE\):
We are given that:
\[ DF \parallel AE \] Therefore, by applying the Basic Proportionality Theorem to \(\Delta ABE\), we get:
\[ \frac{BD}{DA} = \frac{BF}{FE} \quad \text{--- (Equation 2)} \]

• Equate Equation 1 and Equation 2:
Notice that the left-hand side of both equations is the same ratio, \(\frac{BD}{DA}\).
Therefore, the right-hand sides must be equal to each other:
\[ \frac{BF}{FE} = \frac{BE}{EC} \] This completes our geometric proof.


Step 4: Final Answer:
Hence, it is proved that \(\frac{BF}{FE} = \frac{BE}{EC}\).