Question

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

Answer 1

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

Now, take ΔAPQ and ΔABC

∠APQ = ∠ABC (corresponding angles)

∠AQP = ∠ACB (corresponding angles)

∴ ΔAPQ ~ ΔABC (by AA similarity criterion)

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

⇒ BC = 3PQ

Hence Proved