Geometry A circle is inscribed in AABC as shown 14 6 AC in t

Geometry A circle is inscribed in AABC as shown 14 6, AC in the figure above. Suppose that AB 12. Find the length of AM, BN, and Co. and BC (Hint: AM AO: BM e BN: NC e OC)

Solution

The sides AB, BC and AC of the triangle ABC are tangent to the inscribed circle at the points M , N and O respectively. Therefore, we have AM = AO = x (say), BM = BN = y(say) and CO = CN = z(say). Then x+y = 6…(1), y+z = 12…(2) and x + z = 14…(3) . From the 1^st and the 2^nd equations, we have y = 6-x and y = 12-z. Therefore, 6-x = 12-z or, z-x = 12-6 or, z-x = 6..(4). Now, on adding the equations 3 and 4, we get x+z+z-x = 14+6 or, 2z= 20 so that z= 10. Then x = 14-z = 14 – 10 = 4 and y = 12-z = 12-10 = 2.

Thus, AM= 4, BN = 2 and CO = 10.