P and Q are points on sides AB and AC respectively, of ΔABC. If AP = 3 cm, PB = 6 cm, AQ = 5 cm and QC = 10 cm, show that BC = 3 PQ.

Now,
In ΔAPQ and ΔABC,
∠QAP = ∠CAB (common)
(given)
∴ ΔACB ∼ ΔAQP
(∵ ΔACB ∼ ΔAQP)
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⇒ BC = 3PQ
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