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 cuts two sides of a triangle proportionally, it is parallel to the third side.
This is the Converse of the Basic Proportionality Theorem (Thales' Theorem).

Given:
In triangle \( \triangle ABC \), line DE 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 \( DE' \parallel BC \), intersecting AC at point \( E' \).

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

Conclusion:
A line that divides two sides of a triangle proportionally is parallel to the third side.

Hence Proved.