In the given figure, the incircle of △ABC touches the sides AB, BC and CA at the points P, Q, R respectively. Show that
AP+BQ+CR=BP+CQ+AR
=12 (Perimeter of △ABC)
[3 MARKS]
Concept: 1 Mark
Application: 2 Marks
Since the lengths of two tangents drawn from an external point to a circle are equal, we have
AP = AR, BQ = BP, CR = CQ
AP + BQ + CR = AR + BP + CQ ...(i)
Perimeter of △ABC
= AB + BC + CA = AP + BP + BQ + CQ + AR + CR
= (AP + BQ + CR) + (BP + CQ + AR)
= 2(AP + BQ + CR) [Using (i)]
So, 2(AP + BQ + CR) = 2(BP + CQ + AR) = Perimeter of triangle ABC
Hence, AP + BQ + CR= BP + CQ + AR
=12 (Perimeter of △ABC).