Question

A circle touches the side BC of a ∆ABC at P and AB and AC are extended respectively to points Q and R. Prove that AQ is half the perimeter of ∆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

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here is your solution

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.