+0 In triangle $ABC$, let $I$ be the incenter of triangle $ABC$. The line through $I$ parallel to $BC$ intersects $AB$ and $AC$ at $M$ and $N$, respectively. If $AB = 5$, $AC = 5$, and $BC = 8$, then find the area of triangle $AMN$.
+129850 
Triangle ABD is 3-4-5 right with AD = 3 height of ABC
cos(ABD) = 4/5
sin (ABD) = 3/5
sin (ABD / 2) = sin (IBD) = sqrt [(1 - 4/5) / 2 ] = sqrt (1/10)
cos (IBD) = sqrt [ 1 - 1/10 ] = 3 /sqrt10
tan (IBD) = 1/3 = ID / 4
ID = 4/3
AD = 3 - (4/3) = 5/3
Triangles AMN and ABC are similar such that AD / AI = (5/3) / 3 = 5/9
[AMN ] = (5/9)^2 [ ABC ] = (5/9)^2 (1/2) (8)(3) = (25 / 81 ) * 12 = 300 / 81 = 100 / 27
