Question

If a line drawn parallel to one side of triangle intersecting the other two sides in distinct points divides the two sides in the same ratio, then it is parallel to third side. State and prove the converse of the above statement.

Hint:

For proving converse, assume the ratio condition and use contradiction or construction.

Answer 1

Statement:
If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
This is the Converse of the Basic Proportionality Theorem (Thales' Theorem).

Given:
In triangle \( \triangle ABC \), a line intersects sides AB and AC at points D and E respectively, such that:
\[ \frac{AD}{DB} = \frac{AE}{EC} \]
To Prove:
Line \( DE \parallel BC \)

Construction:
Through point D on AB, draw a line \( DE' \) such that \( DE' \parallel BC \), and let it intersect AC at point \( E' \).

Step 1: Apply Basic Proportionality Theorem
In triangle \( \triangle ABC \), since \( DE' \parallel BC \), by the Basic Proportionality Theorem:
\[ \frac{AD}{DB} = \frac{AE'}{E'C} \]
Step 2: Use the given ratio condition
It is already given that: \[ \frac{AD}{DB} = \frac{AE}{EC} \]
From both equations: \[ \frac{AE'}{E'C} = \frac{AE}{EC} \Rightarrow \text{The ratios are equal} \]
Step 3: Conclude using uniqueness of point
Since point E and point E' lie on the same line AC and divide it in the same ratio, the point must be the same:
\[ E \equiv E' \Rightarrow \text{The point E lies on the line } DE' \parallel BC \]
Therefore, \( DE \parallel BC \)

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

Hence Proved.