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 $AMN$.

LiIIiam0216 May 5, 2024

#1**0 **

Calculations: Given:

cos(ABD) = 4/5

sin(ABD) = 3/5

We can calculate:

sin(ABD / 2) = sin(IBD) = sqrt[(1 - 4/5) / 2] = sqrt(1/10)

cos(IBD) = sqrt[1 - 1/10] = 3/sqrt(10)

tan(IBD) = 1/3 = ID / 4

ID = 4/3

AD = 3 - (4/3) = 5/3

Triangles AMN and ABC are similar, and we have: AD / AI = (5/3) / 3 = 5/9

Calculating the areas of the triangles: [AMN] = (5/9)^2 [ABC] = (5/9)^2 (1/2) (8)(3) = (25 / 81 ) * 12 = 300 / 81 = 100 / 27

Answer: 100 / 27

estrellatineoo May 5, 2024