Question

A circle touches the side Bc of triangle ABC at P, and touches AB and BC produced at Q and R show that AQ = HALF perimeter of triangle ABC

Answer 1

Given:  A circle touching the side BC of ΔABC at P and AB, AC produced at Q and R respectively.

RTP: AP = 1/2 (Perimeter of ΔABC)

Proof: Lengths of tangents drawn from an external point to a circle are equal.
          ⇒ AQ = AR, BQ = BP, CP = CR.
          Perimeter of ΔABC = AB + BC + CA
                                      = AB + (BP + PC) + (AR – CR)
                                      = (AB + BQ) + (PC) + (AQ – PC) [AQ = AR, BQ = BP, CP = CR]
                                      = AQ + AQ
                                      = 2AQ

           ⇒ AQ = 1/2 (Perimeter of ΔABC)
 
           ∴ AQ is the half of the perimeter of ΔABC.