+1067 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 $AMN$.
+128732 (1/2)BC = 4
Call the altitude AD = sqrt [ 5^2 - 4^2] = 3
tan ( ABC) = (3/4)
tan (ABC / 2) = 1 -cos (ABC) / sin (ABC) = [ 1 - 4/5 ] / [(3/5 ] = 1/3
Equation of angle bisector of ABC = y =(1/3)x
Intersection of this line with angle bisector BAC {line x = 4 }
y= (1/3) (4) = 4/3
Height of triangle AMN = 3 - 4/3 = 5/3
Area of [ABC] = (1/2) AD * BC = (1/2) (3) (8) = 12
Triangle AMN similar to triangle ABC
Scale factor of AMN/ ABC = (5/3) / 3 = 5/9
[ AMN ] = 12 * [ (5/9 ] ^2 = 100 / 27 ≈ 3.7
