Question

A circle is touching the side BC of triangle ABC at P and touching AB and AC produced Q and R respectively . Prove that AQ = 1/2[Perimeter of triangle ABC.

Answer 1

this is the answer of yr question

Answer 2

Answer:

: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.

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