Question

In the figure, $DE \parallel AC$ and $DF \parallel AE$. Prove that $\dfrac{BF}{FE} = \dfrac{BE}{EC}$.

Hint:

When parallels are drawn in triangles, always apply the Basic Proportionality Theorem (BPT) in two different triangles and equate the common ratios to establish the required relation.

Answer 1

 

Step 1: Apply BPT in $\triangle ABC$.
Since $DE \parallel AC$, by Basic Proportionality Theorem (BPT): \[ \frac{BD}{BA} = \frac{BE}{BC}. (1) \]

Step 2: Apply BPT in $\triangle ABE$.
Since $DF \parallel AE$: \[ \frac{BD}{BA} = \frac{BF}{BE}. (2) \]

Step 3: Equating ratios.
From (1) and (2): \[ \frac{BF}{BE} = \frac{BE}{BC} \;\Rightarrow\; BF = \frac{BE^2}{BC}. \]

Step 4: Express $FE$.
$FE = BE - BF = BE - \frac{BE^2}{BC} = BE\left(1 - \frac{BE}{BC}\right) = BE \cdot \frac{BC - BE}{BC} = BE \cdot \frac{EC}{BC}$.

Step 5: Required ratio.
\[ \frac{BF}{FE} = \frac{\tfrac{BE^2}{BC}}{\tfrac{BE \cdot EC}{BC}} = \frac{BE}{EC}. \] \[ \boxed{\dfrac{BF}{FE} = \dfrac{BE}{EC}} \]