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

This completes the proof.