Question

a circle touches the side BC of triangle ABC at P and the extended sides AB and AC at Q and R respectively . prove that AQ=1/2 (BC+CA+AB)

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.

Answer 2

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