Question

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step by step explaination
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Answer 1

Sry dint have a paper rn so couldn't draw the diag but the proof might help!

Given: A circle touching the side BC of ΔABC at P

AB, AC produced at M and N respectively.

RTP: AM = 1/2 (Perimeter of ΔABC)

Proof: Lengths of tangents drawn from an external point to a circle are equal.

AM = AM,

BM = BP

CP = CN

Perimeter of ΔABC = AB+BC+CA = AB + (BP + PC) + (AN– CN)

= (AB + BM) + (PC) + (AM – PC) [AM= AN, BM = BP, CP = CN] = 2AM

⇒ AM= 1/2 (Perimeter of ΔABC) ∴