P and Q are respectively the points on the sides AB and AC of a ΔABC. If AP = 2 cm, PB = 6 cm, AQ = 3 cm and QC = 9, Prove that BC = 4PQ.

Here,

and

∴ PQ || BC [by the converse of basic proportionality theorem]

Now, take APQ and ABC

∠APQ = ∠ABC (corresponding angles)

∠AQP = ∠ACB (corresponding angles)

∴ APQ ~ ABC (by AA similarity criterion)

Since, triangles are similar, hence corresponding sides will be proportional

⇒BC = 4PQ

Hence Proved