Question

ABC is a triangle. A circle touches sides AB and AC produced and side BC at X,Y and Z respectively.Show that AX=1/2 perimeter of triangle ABC.

Answer 1

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.