Question

prove that ar (ABC) = half radius *perimeter of triangle ABC

Answer 1

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.