Question

Prove that the line dividing any two sides of a triangle in a same ratio is parallel to the third side of the triangle.

Hint:

The proof by contradiction is a powerful tool in geometry. The strategy is to assume the opposite of what you want to prove and show that this assumption leads to a logical impossibility.

Answer 1

Step 1: Understanding the Concept:
This is the statement of the Converse of the Basic Proportionality Theorem (also known as Thales's Theorem or the Intercept Theorem). We need to prove this geometric property using a proof by contradiction.
Step 2: Key Formula or Approach:
We will use the Basic Proportionality Theorem (BPT), which states that if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
Step 3: Detailed Explanation:
Given: A triangle \(\triangle ABC\) and a line DE intersecting sides AB and AC at points D and E respectively, such that \( \frac{AD}{DB} = \frac{AE}{EC} \).
To Prove: DE is parallel to BC (DE || BC).
Proof by Contradiction:
Let us assume that DE is not parallel to BC.
If DE is not parallel to BC, then there must be another line through D which is parallel to BC.
Let's draw a line DF such that DF || BC, where F is a point on the side AC.
Since DF || BC, by the Basic Proportionality Theorem (BPT), we have:
\[ \frac{AD}{DB} = \frac{AF}{FC} \quad \ldots (i) \] But, it is given that:
\[ \frac{AD}{DB} = \frac{AE}{EC} \quad \ldots (ii) \] From equations (i) and (ii), we can conclude:
\[ \frac{AF}{FC} = \frac{AE}{EC} \] Adding 1 to both sides of the equation:
\[ \frac{AF}{FC} + 1 = \frac{AE}{EC} + 1 \] \[ \frac{AF + FC}{FC} = \frac{AE + EC}{EC} \] Since AF + FC = AC and AE + EC = AC, we get:
\[ \frac{AC}{FC} = \frac{AC}{EC} \] This implies that \( FC = EC \).
This is only possible when the points F and E coincide, meaning they are the same point.
Therefore, the line DF is the same as the line DE.
Since we constructed DF || BC, it must be that DE || BC.
This contradicts our initial assumption. Hence, our assumption was wrong.
Thus, the line DE is parallel to BC.
\[ \text{Hence Proved.} \]