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Turkey NMO 2007 1st Round - P25 (Geometry)

Source:

October 5, 2012
geometry

Problem Statement

Let A,B,CA, B, C be points on a unit circle such that ∣AB∣=∣BC∣|AB|=|BC| and m(ABC^)=72∘m(\widehat{ABC})=72^\circ. Let DD be a point such that △BCD\triangle BCD is equilateral. If ADAD meets the circle at DD, what is ∣DE∣|DE|?
<spanclass=′latex−bold′>(A)</span> 12<spanclass=′latex−bold′>(B)</span> 32<spanclass=′latex−bold′>(C)</span> 22<spanclass=′latex−bold′>(D)</span> 3−1<spanclass=′latex−bold′>(E)</span> None of the above <span class='latex-bold'>(A)</span>\ \dfrac 12 \qquad<span class='latex-bold'>(B)</span>\ \dfrac {\sqrt 3}2 \qquad<span class='latex-bold'>(C)</span>\ \dfrac {\sqrt 2}2 \qquad<span class='latex-bold'>(D)</span>\ \sqrt 3 -1 \qquad<span class='latex-bold'>(E)</span>\ \text{None of the above}
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