Question

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

Hint:

To prove parallel lines using proportional sides, apply the Converse of the Basic Proportionality Theorem.

Answer 1

Converse Statement: If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side. Given: In \( \triangle ABC \), a line intersects \( AB \) and \( AC \) at points \( D \) and \( E \) respectively such that \[ \frac{AD}{DB} = \frac{AE}{EC} \] To Prove: \( DE \parallel BC \) Construction: Draw a line \( D'E' \parallel BC \) intersecting \( AB \) at \( D' \) and \( AC \) at \( E' \). Proof: By Basic Proportionality Theorem: \[ \frac{AD'}{D'B} = \frac{AE'}{E'C} \] But it’s given that: \[ \frac{AD}{DB} = \frac{AE}{EC} \] So, by uniqueness of ratio: \[ D = D',\ E = E' \Rightarrow DE \parallel BC \] \[ \boxed{\text{Hence proved: } DE \parallel BC} \]