Converse Of Basic Proportionality Theorem

Theorem: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

Prerequisites:
Basic Proportionality Theorem (proof)
Unique parallel through a point (proof)

proof:

Let there be a $\triangle ABC$ and a line $l$ intersecting the sides AB and AC at the points D and E respectively, as shown in the figure, such that:

$\qquad\quad\dfrac{AD}{DB} = \dfrac{AE}{EC}$

We need to show that $l \parallel BC$. For this, let us assume that $l \not\parallel BC$. Then there must exist a unique line DF through the point D, such that DF $\parallel$ BC.

Since DF $\parallel$ BC, hence by Basic Proportionality Theorem, we get:

$\qquad\quad\;\:\!\dfrac{AD}{DB} = \dfrac{AF}{FC}\\[12pt]
\text{But, }\quad\dfrac{AD}{DB} = \dfrac{AE}{EC}\qquad\qquad\qquad\qquad\text{(given)}\\[12pt]
\therefore\quad\quad\:\dfrac{AF}{FC} = \dfrac{AE}{EC}\\[12pt]
\Rightarrow\quad\;\;\;\dfrac{AF}{FC} +1 = \dfrac{AE}{EC} +1\\[12pt]
\Rightarrow\quad\;\;\;\dfrac{AF+FC}{FC} = \dfrac{AE+EC}{EC}\\[12pt]
\Rightarrow\quad\;\;\;\:\dfrac{AC}{FC} = \dfrac{AC}{EC}\\[12pt]
\Rightarrow\qquad\; FC = EC$

But this is a contradiction as F $\neq$ E. Hence, our assumption was false. Therefore, $l \parallel BC$.

Q.E.D.


Recommended:
Angle Bisector Theorem
Midpoint Theorem
Location of centroid