Area of triangle ABC
Find the semi-perimeter, S = ( 8 + 6+ 4) /2 = 9
Area = sqrt [ 9 (9-8) (9-6) (9-4) ] = sqrt [ 9 * 3 * 5 ] = 3sqrt (15)
The circle is an inscribed circle in a triangle....its radius, R, can be found as
Area / S = R
3sqrt (15) / 9 = R = sqrt (15) / 3
And the height of triangle ABC can be found as
3sqrt (15) = (1/2) BC * height
6sqrt (15) = 4 * height
(3/2) sqrt (15) = height of ABC
Height of triangle AMN = (3/2)sqrt (15) - 2R = (3/2)sqrt (15) - (2/3)sqrt (15) = (5/6)sqrt (15)
And triangle AMN is similar to triangle ABC....so....
MN/ height of AMN = BC /height of ABC
MN/ (5/6)sqrt (3) = 4 / [(3/2)sqrt (15)]
MN / (5/6) = 4 /(3/2)
MN = 4 (5/6) / (3/2)
MN = (4)(2/3)(5/6) = 40/18 = 20/9