Statement (Converse of BPT / Theorem 6.2): If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Given: In △ABC, DE divides AB and AC such that $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
To Prove: DE ∥ BC
Proof: Assume DE is not parallel to BC. Draw DE′ ∥ BC.
Since DE′ ∥ BC, by Basic Proportionality Theorem (Theorem 6.1):
$$\frac{AD}{DB} = \frac{AE'}{E'C} \quad \cdots (1)$$
But it is given that:
$$\frac{AD}{DB} = \frac{AE}{EC} \quad \cdots (2)$$
From (1) and (2):
$$\frac{AE}{EC} = \frac{AE'}{E'C}$$
Adding 1 to both sides:
$$\frac{AE}{EC} + 1 = \frac{AE'}{E'C} + 1$$
$$\frac{AE + EC}{EC} = \frac{AE' + E'C}{E'C}$$
$$\frac{AC}{EC} = \frac{AC}{E'C}$$
Therefore, EC = E′C, which means E and E′ coincide.
Hence, DE ∥ BC. [Proved]
Source: Chapter 6, Section 6.3 — Theorem 6.2
---