Question

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

Answer 1

We have : 

AP/AB = 2/6 = 1/3 and AQ/AC = 3/9 = 1/3 

⇒ AP/AB = AQ/AC 

In ∆ APQ and ∆ ABC, we have: 

AP/AB = AQ/AC

∠A = ∠A

Therefore, by AA similarity theorem, we get: 

∆ APQ - ∆ ABC 

Hence, PQ/BC = AQ/AC = 1/3 

⇒ PQ/BC = 1/3 

⇒ BC = 3PQ

This completes the proof.