Question

Prove that, "If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion."

Hint:

The basic proportionality theorem (Thales' theorem) is a fundamental result for parallel lines in triangles.

Answer 1

Step 1: Consider \( \triangle ABC \) with a line \( DE \parallel BC \) intersecting \( AB \) at \( D \) and \( AC \) at \( E \).
Step 2: In \( \triangle ADE \) and \( \triangle ABC \), since \( DE \parallel BC \), corresponding angles are equal: \[ \angle ADE = \angle ABC, \quad \angle DEA = \angle BCA. \] Step 3: By AA similarity criterion: \[ \triangle ADE \sim \triangle ABC. \] Step 4: From the similarity property: \[ \frac{AD}{DB} = \frac{AE}{EC}. \] Step 5: This proves that the line \( DE \) divides the sides \( AB \) and \( AC \) in the same proportion.