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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.
Solution
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.
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