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 = 3PQ.

There are two triangles here, ∆APQ and ∆ABC. We shall prove these triangles to be similar.


&



Also, A = A [common angle]


So by AA-similarity criteria,


∆APQ ∆ABC


Thus,



And we know


BC = 3×PQ


Hence, proved.


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