a circle touches the side BC of triangle abc at P and touch a b and ac produced at q and R respectively prove that PQ equal to half perimeter of triangle ABC
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Step-by-step explanation:
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.
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