Question

A circle is touching the side BC of triangle ABC at p and touching a b and ac produced at q and r respectively prove that AQ=1/2(perimeter of ABC)..

Answer 1

Answer:

Step-by-step explanation:

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.