+1507 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 $AMN$.
+129845 The triangle is isosceles
Let B = (0,0) C =(8,0) A = (3,4)
Height of triangle ABC = sqrt [5^2 -4^2] = sqrt [9] = 3
Area of ABC = (1/2) (8)(3) = (1/2) BC * height = 12
The incenter = (4, y)
To find y
[ 3* 8 + 0 * 5 + 0 * 5 ] / [ 5 + 5 + 8 ] = 24 / 18 = 4/3
Height of AMN = 3 - 4/3 = 5/3
AMN similar to ABC
Scale factor of AMN to ABC = [5/3]/ 3 = 5/9
[AMN ] = (5/9)^2 [ ABC ] = (25/81) [ 12] = 100 / 27
