+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 $AMN$.
+130081 
I = (4,r) where r is the radius of the incircle
To find r
[ ABC] = (1/2)(8 + 5 + 5) r
(1/2) 8 * 3 = (1/2) (18) r
r = 4/3
l = (4, 4/3)
ID = 4/3
AI = 3 - (4/3) = 5/3
Triangles AMN and ABC are similar such that AI / AD = = (5/3) / 3 = 5/9
[AMN ] = (5/9)^2 [ ABC ] = (5/9)^2 (1/2) (8)(3) = (25 / 81 ) * 12 = 300 / 81 = 100 / 27
