Question 4

We will use the Basic Proportionality Theorem (Thales's Theorem) to solve this problem.

Step 1 — Apply BPT in $\triangle ABC$

Let's look at $\triangle ABC$. We are given that DE || AC. By the Basic Proportionality Theorem, the sides are divided proportionally.

$\frac{\text{BD}}{\text{DA}} = \frac{\text{BE}}{\text{EC}}$

$\boxed{\frac{\text{BD}}{\text{DA}} = \frac{\text{BE}}{\text{EC}} \quad \text{(1)}}$

Step 2 — Apply BPT in $\triangle BAE$

Now, let's consider $\triangle BAE$. We are given that DF || AE. By the Basic Proportionality Theorem, the sides are divided proportionally.

$\frac{\text{BD}}{\text{DA}} = \frac{\text{BF}}{\text{FE}}$

$\boxed{\frac{\text{BD}}{\text{DA}} = \frac{\text{BF}}{\text{FE}} \quad \text{(2)}}$

Step 3 — Equate the ratios

From equation (1), we know $\frac{\text{BD}}{\text{DA}} = \frac{\text{BE}}{\text{EC}}$. From equation (2), we know $\frac{\text{BD}}{\text{DA}} = \frac{\text{BF}}{\text{FE}}$. Both ratios are equal to $\frac{\text{BD}}{\text{DA}}$. Therefore, they must be equal to each other.

$\frac{\text{BF}}{\text{FE}} = \frac{\text{BE}}{\text{EC}}$

Answer

(i) We have proved that $\frac{\text{BF}}{\text{FE}} = \frac{\text{BE}}{\text{EC}}$.