Asked in: March 2024 board paper (the basic proportionality theorem is the most important proof of this chapter)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 s…

Answer: Proof of the basic proportionality theorem: \(\dfrac{AP}{PB} = \dfrac{AQ}{QC}\)

[Figure: \(\Delta ABC\); line \(l\) parallel to BC cuts AB at P and AC at Q. Seg PC and seg BQ are drawn.]

Given: In \(\Delta ABC\), line \(l\) || side BC and line \(l\) intersects side AB at P and side AC at Q.

To prove: \(\dfrac{AP}{PB} = \dfrac{AQ}{QC}\).

Construction: Draw seg PC and seg BQ.

Proof: \(\Delta APQ\) and \(\Delta PQB\) have the same height (the perpendicular from Q to AB). \(\therefore \dfrac{A(\Delta APQ)}{A(\Delta PQB)} = \dfrac{AP}{PB}\) ...... (I) (areas proportional to bases)

\(\Delta APQ\) and \(\Delta PQC\) have the same height (the perpendicular from P to AC). \(\therefore \dfrac{A(\Delta APQ)}{A(\Delta PQC)} = \dfrac{AQ}{QC}\) ...... (II)

\(\Delta PQB\) and \(\Delta PQC\) have the same base PQ and lie between the same parallel lines PQ and BC, so their heights are equal. \(\therefore A(\Delta PQB) = A(\Delta PQC)\) ...... (III)

From (I), (II) and (III), \(\dfrac{A(\Delta APQ)}{A(\Delta PQB)} = \dfrac{A(\Delta APQ)}{A(\Delta PQC)}\), hence \(\dfrac{AP}{PB} = \dfrac{AQ}{QC}\). Hence proved.