Q10 of 186 Page 211

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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