Q15 of 95 Page 6

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


Here,


and



∴ PQ || BC [by converse of basic proportionality theorem]


Now, take APQ and ABC


∠APQ = ∠ABC (corresponding angles)


∠AQP = ∠ACB (corresponding angles)


∴ APQ ~ ABC (by AA similarity criterion)


Since, triangles are similar, hence corresponding sides will be proportional





⇒BC = 3PQ


Hence Proved


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