# 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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