Question

A circle touch the side BC of ∆ABC at P and touches AB and AC produced at Q and R respectively. Prove that AQ=1/2 (perimeter of ∆ABC).

Answer 1

Answer:

Step-by-step explanation:

Sol:

Given:  A circle touching the side BC of ΔABC at P and AB, AC produced at Q and R respectively.

RTP: AQ = 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.