Question

in the figure the side Bc of triangle ABC touches a circle with centre O at R. prove that Ap=1/2 perimeter(Abc)​

Answer 1

Step-by-step explanation:

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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