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Let two circles S_{1} and S_{2} meet at the points A and B. A line through A meets S_{1} again at C and S_{2} again at D. Let M, N, K be three points on the line segments CD, BC, BD respectively, with MN parallel to BD and MK parallel to BC. Let E and F be points on those arcs BC of S_{1} and BD of S_{2} respectively that do not contain A. Given that EN is perpendicular to BC and FK is perpendicular to BD prove that \angle EMF=90^{\circ}.

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