Question

step by step ans this q​

Answer 1

Answer:

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.

Answer 2

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