asw the q
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u don't know ans don't write.
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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) ∴
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