a circle touches the side BC of a triangle ABC at P and AB and AC are extended respectively to points q and R prove that AQ is half the perimeter of triangle ABC
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Figure:It is given in the attachment below.
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 [AB+BQ=AQ and PC-PC=0]
Perimeter of ΔABC = 2AQ
AQ = 1/2 (Perimeter of ΔABC)
∴ AQ is the half of the perimeter of ΔABC.
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