Question

Show that perimeter of∆ABC=2(ab+ bq+ cr)

Answer 1

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

Sol:

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