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)

Answer 1

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