Question

In the given figure, the incircle of $△ABC$ touches the sides AB, BC and CA at the points P, Q, R respectively. Show that
$AP+BQ+CR=BP+CQ+AR$
$=\frac{1}{2}(Perimeterof△ABC)$
[3 MARKS]

Answer 1

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)

$Perimeterof△ABC$

= 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
$=\frac{1}{2}(Perimeterof△ABC)$.