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Question
A circle touches side BC at point P of the ΔABC, from outside of the triangle. Further extended lines AC and AB are tangents to the circle at N and M respectively. Prove that : AM = `1/2` (Perimeter of ΔABC)
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Solution
Since two tangents drawn from an external point to a circle are equal, we establish the following equalities:
- AM = AN
- BP = BM
- CP = CN
The perimeter of △ABC is: AB + BC + CA
Substituting in terms of tangents: (AM + BM) + (BP + PC) + (CN + AN)
AM + AN + (BM + BP) + (CN + CP)
AM + AN + BM + CN + BP + CP
Since BM = BP and CN = CP, we simplify to:
AM + AN + BP + BM + CP + CN = 2(AM + BP + CP)
Since BP + CP = BC, we get:
`AM + AN = 1/2 xx` Perimeter of △ABC
`AM = 1/2 xx` Perimeter of △ABC
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