Problem:
Equilateral triangle ABC has side length 840. Point D lies on the same side of line BC as A such that BD⊥BC. The line ℓ through D parallel to line BC intersects sides AB and AC at points E and F, respectively. Point G lies on ℓ such that F is between E and G,△AFG is isosceles, and the ratio of the area of △AFG to the area of △BED is 8:9. Find AF.
Solution:
Using the fact that △ABC is equilateral, we can angle chase and find that triangle AGF is a 30∘–30∘–120∘ triangle, and triangle BED is a 30∘–60∘–90∘ triangle.
Set AF=x. Since ∠AGF=120∘ and triangle AGF is isosceles with ∠GAF=∠GFA=30∘, we also have FG=x.
Given that FC=840−x, we have EB=FC=840−x.
In triangle BED, by the 30∘–60∘–90∘ ratio, we get DE=2840−x and DB=2840−x⋅3.
We calculate the area of triangle AGF using the formula for area with sine:
