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 $AMN$.

Stanry May 25, 2024

#1**0 **

To find the area of triangle AMN, we can use the fact that the area of a triangle is equal to half the product of its base and height. In this case, we need to find the height of triangle AMN with respect to base MN.

1. Let D be the point where the line through I parallel to BC intersects MN.

2. Since I is the incenter of triangle ABC, we know that ID is perpendicular to MN.

3. Therefore, triangle IDM is a right triangle with ID as its height and MD as its base.

4. By properties of parallel lines, triangles ABC and AMN are similar.

5. This means that the ratio of corresponding side lengths in these triangles is equal. So, AB/AM = BC/MN.

6. Since AB = 5 and BC = 8, we can find MN as (5/8) × MN = 5, which gives MN = 40/5 = 8.

7. Now, ID is the height of triangle IDM, and MD = MN = 8.

8. Using the Pythagorean theorem in triangle IDM, we can find ID as ID^2 + 4^2 = 5^2, which gives ID = 3.

9. The area of triangle AMN is then (1/2) × MD × ID = (1/2) × 8 × 3 = 12. Therefore, the area of triangle AMN is 12

Hi69420 May 25, 2024