Using the figure given below, prove that AR = 1/2 (perimeter of triangle ABC)

Clearly, in the given figure

Perimeter of ΔABC = AC + BC + AB …[1]

As we know that, Tangents drawn from an external point to a circle are equal.

So we have

CR = CP [1] [Tangents from point C]

PB = BQ [2] [Tangents from point B]

Now Perimeter of Triangle ABC

= AC + BC + AB

= AC + CP + PB + AB

= AC + CR + BQ + AB [From 1 and 2]

= AR + AQ

Now,

AQ = AR as tangents drawn from an external point to a circle are equal

So we have

Perimeter of ΔABC = AR + AR = 2AR