Question

In the given figure, prove that: BP + CQ + AR = `1/2` × perimeter of ΔABC.

Answer 1

Let the circle touch,

BC, CA, AB at points P, Q, R, respectively.

Since tangents drawn from an external point to a circle are equal:

From vertex A: AQ = AR

From vertex B: BP = BQ

From vertex C: CP = CR

Consider the sum AR + BP + CQ.

Using equal tangents, rewrite as: AR + BP + CQ = AQ + BQ + CR

Group the terms as: AR + BP + CQ = (AB − BQ) + (BC − CP) + (AC - CR)

Use equal tangents to substitute: BQ = BP, CP = CR, AQ = AR

Rearranging terms leads to: AR + BP + CQ = `(AB + BC + AC)/2`

Since the perimeter of triangle ABC is AB + BC + AC, we have:

`BP + CQ + AR = 1/2` × perimeter of ΔABC.