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.[Figure: Δ ABC with A at the top, B at the lower left and C at the lower right; a li…

Answer: Proof of the basic proportionality theorem (below).

Given: In Δ ABC, line PQ ∥ side BC; line PQ 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: Δ APQ and Δ PQB have the same height (the perpendicular distance from Q to line AB).

\(\therefore \dfrac{A(\Delta APQ)}{A(\Delta PQB)} = \dfrac{AP}{PB}\) ...(I) (areas proportional to bases)

Δ APQ and Δ PQC have the same height (the perpendicular distance from P to line AC).

\(\therefore \dfrac{A(\Delta APQ)}{A(\Delta PQC)} = \dfrac{AQ}{QC}\) ...(II)

Δ PQB and Δ PQC have the same base PQ and lie between the same parallel lines PQ and BC, so they have equal heights.

\(\therefore A(\Delta PQB) = A(\Delta PQC)\) ...(III)

From (I), (II) and (III): \(\dfrac{AP}{PB} = \dfrac{AQ}{QC}\).