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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below to enlarge it.
Let BD and A₁C₂ intersect at P. Join A₁B and BC₂. Then A₁BC₂D is a rectangle. Thus PA₁=PB=PC₂=PD, and so A₁C₂ is the common tangent of O₁ and O₂. ***Clearly, A₁B//A₃O₂//DC, and BO₂=O₂C, so A₁A₃=A₃D. Similarly, C₂C₄=C₄D. Therefore, A₁C₂//A₃C₄ and A₁C₂=2×A₃C₄. Since A₁BC₂D is rectangle, thus A₁C₂=BD=2×A₃C₄. Also since P is mid-point of A₁C₂, so M is mid-point of A₃C₄. ***Since A₁O₁//A₃B//DO₄, and also A₁C₂//A₃C₄, so A₃C₄ is tangent to circle O₂. Similarly A₃C₄ is tangent to circle O₄. And hence A₃C₄ is the common tangent of O₂ and O₄. Since BD passes through M, and M is mid-point of A₃C₄, thus M lies on the radical axis of O₂ and O₄, which is BD. Therefore BD passes through E, which also lies on the radical axis of O₂ and O₄. Hence, B,E,D are collinear.