In figure P, Q are points on sides AB, AC respectively of ∆ABC such that ar (BCQ) = ar (BCP). Then,
PA || BC
BP || QC
PQ || BC
AB || CP
Since the triangles ∆BCQ and ∆BCP are on the same base and their areas are equal, therefore they lie between two parallels.
In figure P, Q are points on sides AB and AC respectively of ∆ABC such that ar (BCQ) = ar (BCP). Then,
In a Δ ABC, P and Q are points on sides AB and AC respectively, such that PQ || BC. If 2.4 cm, AQ =2 cm, QC = 3cm and BC =6cm, Finnd AB and PQ.