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HMMT 2004 U31

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PP is a point inside triangle ABCA B C, and lines AP,BP,CPA P, B P, C P intersect the opposite sides BC,CA,ABB C, C A, A B in points D,E,FD, E, F, respectively. It is given that ∠APB=90∘\angle A P B=90^{\circ}, and that AC=BCA C=B C and AB=BDA B=B D. We also know that BF=1B F=1, and that BC=999B C=999. Find AFA F.

Dataset text NuminaMath olympiads. Quoted with source.

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\frac{499}{500}

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