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## Editorial

Let $A F=x \Longrightarrow B E=840-x$.

Draw altitude from $F$ to $A G$. Can you spot the similar triangles?

Equilateral triangle $ABC$ has side length $840$. Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$. The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$, respectively. Point $G$ lies on $\ell$ such that $F$ is between $E$ and $G$, $\triangle AFG$ is isosceles, and the ratio of the area of $\triangle AFG$ to the area of $\triangle BED$ is $8:9$. Find $AF$.

Geometry

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Need a hint? Checkout the editorial.

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Let $A F=x \Longrightarrow B E=840-x$.

Draw altitude from $F$ to $A G$. Can you spot the similar triangles?