Tangents AB, BC, AC to circle O at points M, N, and P, Respectively AB= 14, BC= 16, AC= 12

Question

find AM, PC, BN.

Answer 1

The incircle :

The incircle is tangent to each of the three sides.AB, BC, and AC.(without extension).

Its center, the incenter I, is the intersection of the bisectors of the three angles.

The inradius r is related to the area (1 / 2)S by, S = (a + b + c )r.

If the incircle is tangent to the sides AB at M, BC at N, and AC at P, then

BM = BN = ( b + c - a ) / 2,

PC = CN = ( c + a - b ) / 2, and

AM = AP = ( a + b - c ) / 2.

Given :

AB = 14, BC = 16, and AC = 12.

Let us consider that, AC = a, AB = b, and BC = c.

To find AM = ( a + b - c ) / 2.

Substitute the values a = 12, b = 14, and c = 16.

AM = (12 + 14 - 16) / 2 = (26 - 16)  / 2 = 10 / 2 = 5.

To find PC = ( c + a - b ) / 2.

PC = (16 + 12 - 14) / 2 = (28 - 14) / 2 = 14 / 2 = 7.

To find BN = ( b + c - a) / 2.

BN = (14 + 16 - 12) / 2 = (30 - 12) / 2 = 18 / 2 = 9.