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: [AGF]=21⋅AF⋅FG⋅sin(∠AFG)=21⋅x⋅x⋅sin(120∘)=21⋅x2⋅23=43x2.
Now for triangle BED: [BED]=21⋅DE⋅DB=21⋅2840−x⋅(2840−x⋅3)=83(840−x)2.
Now set up the given area ratio: [BED][AGF]=83(840−x)243x2=4(840−x)28x2=(840−x)22x2.
We are told this ratio equals 98, so: (840−x)22x2=98.
Divide both sides by 2: (840−x)2x2=94.
Taking the positive square root (since x<840), we get: 840−xx=32.
Cross-multiplying: 3x=2(840−x), so 3x=1680−2x, and therefore 5x=1680, giving x=336.
