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: Set up two pairs of similar triangles.
Since $DF \parallel AE$, triangles $BDF$ and $BAE$ are similar (AA), giving $\frac{BF}{BE} = \frac{BD}{BA}$ ... (i)
Since $DE \parallel AC$, triangles $BDE$ and $BAC$ are similar (AA), giving $\frac{BE}{BC} = \frac{BD}{BA}$ ... (ii)
Step 2: Combine the two ratios.
From (i) and (ii), since both equal $\frac{BD}{BA}$:
\[ \frac{BF}{BE} = \frac{BE}{BC} \]
Step 3: Apply componendo-dividendo.
Subtracting each side from $1$: $1 - \frac{BF}{BE} = 1 - \frac{BE}{BC}$, which gives $\frac{BE - BF}{BE} = \frac{BC - BE}{BC}$, i.e. $\frac{FE}{BE} = \frac{EC}{BC}$.
Taking reciprocals and cross-multiplying with the original ratio $\frac{BF}{BE} = \frac{BE}{BC}$ leads back to:
\[ \frac{BF}{FE} = \frac{BE}{EC} \]
Hence the required result is proved using similar triangles and componendo-dividendo, a route different from a direct double application of the Basic Proportionality Theorem.