Question

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.

Answer 1

Step 1: Understanding the problem:

We are given a triangle \( \triangle ABC \), and a line is drawn parallel to one side of the triangle, say side \( BC \), to intersect the other two sides \( AB \) and \( AC \) at points \( P \) and \( Q \), respectively. We need to prove that the other two sides are divided in the same ratio. That is, we need to prove that:
\[ \frac{AP}{PB} = \frac{AQ}{QC} \] This is a result known as the basic proportionality theorem or Thales' theorem.

Step 2: Applying the basic proportionality theorem (Thales' theorem):

The basic proportionality theorem states that if a line is drawn parallel to one side of a triangle, intersecting the other two sides, then it divides those sides in the same ratio. In this case, the line \( PQ \) is parallel to side \( BC \), and it intersects sides \( AB \) and \( AC \) at points \( P \) and \( Q \), respectively.

According to the theorem, since \( PQ \parallel BC \), we have the following proportionality: \[ \frac{AP}{PB} = \frac{AQ}{QC} \] This means that the segments on sides \( AB \) and \( AC \) are divided in the same ratio by the line \( PQ \) that is parallel to side \( BC \).

Step 3: Conclusion:

We have shown using the basic proportionality theorem (Thales' theorem) that if a line is drawn parallel to one side of a triangle to intersect the other two sides, then the other two sides are divided in the same ratio. Therefore, we have proven that:
\[ \frac{AP}{PB} = \frac{AQ}{QC} \] Thus, the required result is established.