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.