Question

A circle is touching the side BC of triangle ABC at P and touching AB and AC producced at Q and R respectively.Prove that AO=1/2 (perimeter ABC).

Answer 1

GIVEN:  A circle touching the side BC of ΔABC at P and AB, AC produced at Q and R.
To Prove :  AQ = 1/2 (Perimeter of ΔABC)
PROOF:
AQ = AR         [From A].........(1)
BQ = BP          [From B].........(2)
CP = CR.          [From C]........(3)
[Lengths of tangents drawn from an external point to a circle are equal.]
Perimeter of ΔABC = AB + BC + CA
Perimeter of ΔABC =AB+ (BP + PC) + (AR - CR)
Perimeter of ΔABC = (AB + BQ) + (PC) + (AQ - PC) 
[From eq 1,2 & 3 , AQ = AR, BQ = BP, CP = CR]
Perimeter of ΔABC = AQ + AQ = 2AQ
Perimeter of ΔABC = 2AQ

AQ = 1/2 (Perimeter of ΔABC)
Hence,  AQ is half the perimeter of ΔABC.

HOPE THIS WILL HELP YOU...

Answer 2

Answer:

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.

Go through with attachment

Please mark me as brainlist.