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B'M perp to A'C' - Iran NMO 2001 (Second Round) - Problem2

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October 4, 2010
geometry proposedgeometry

Problem Statement

Let ABCABC be an acute triangle. We draw 33 triangles B′AC,C′AB,A′BCB'AC,C'AB,A'BC on the sides of ΔABC\Delta ABC at the out sides such that: ∠B′AC=∠C′BA=∠A′BC=30∘   ,   ∠B′CA=∠C′AB=∠A′CB=60∘ \angle{B'AC}=\angle{C'BA}=\angle{A'BC}=30^{\circ} \ \ \ , \ \ \ \angle{B'CA}=\angle{C'AB}=\angle{A'CB}=60^{\circ} If MM is the midpoint of side BCBC, prove that B′MB'M is perpendicular to A′C′A'C'.
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