olymmath

HMMT 2014 U14

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Let ABCDA B C D be a trapezoid with AB∥CDA B \| C D and ∠D=90∘\angle D=90^{\circ}. Suppose that there is a point EE on CDC D such that AE=BEA E=B E and that triangles AEDA E D and CEBC E B are similar, but not congruent. Given that CDAB=2014\frac{C D}{A B}=2014, find BCAD\frac{B C}{A D}.

Dataset text NuminaMath olympiads. Quoted with source.

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\sqrt{4027}

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