Question

triangle ABC a circle touches sides ab and ac produced and side BC at x y z respectively show that X is equals to half the perimeter of triangle 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.

To prove: 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.