Question

P and Q are points on the sides AB and AC respectively of a ΔABC. 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 Δ APQ and Δ ABC, we have:  

`(AP)/(AB)=(AQ)/(AC)` 

∠𝐴= ∠𝐴
Therefore, by AA similarity theorem, we get: 

Δ APQ - Δ ABC 

Hence,` (PQ)/(BC)=(AQ)/(AC)=1/3` 

⇒` (PQ)/(BC)=1/3`

⟹ BC = 3PQ  

This completes the proof.