Simson Line in DisguiseThe figure below shows a triangle ABC with a point D. DE is perpendicular to DC (E is on BA extended), DF is perpendicular to DA (F is on BC extended), and DG is perpendicular to DB (G is on AC extended). Prove that points E, F, G are collinear.
 |Activities| |Contact| |Front page| |Contents| |Geometry| |Store| Copyright © 1996-2012 Alexander BogomolnyThe figure below shows a triangle ABC with a point D. DE is perpendicular to DC (E is on BA extended), DF is perpendicular to DA (F is on BC extended), and DG is perpendicular to DB (G is on AC extended). Prove that points E, F, G are collinear.
SolutionLet C(Δ) denote the circumcircle of triangle Δ. Draw C(DEF), C(DFG), and C(DGE). Let their centers be C', A', and B', respectively. Consider a pair of them, say C(DFG), and C(DGE). The two circles meet at two points D and G. Therefore, their center line B'C' is perpendicular to the common chord DG and passes through the midpoint Gm of the latter. Similar considerations apply to the two remaining pairs of the circles. At this point, I wish to strengthen the statement. From the forgoing observations, the sides of ΔA'B'C' are parallel to the cevians AD, BD, and CD. Which brings to mind the Maxwell theorem. According to that theorem, the cevians in ΔA'B'C' parallel to the sides of ΔABC are concurrent. Let's denote the point of concurrency D'. Then we have The four points A, F, G, and D' are concurrent. ... to be continued ... |Activities| |Contact| |Front page| |Contents| |Geometry| |Store| Copyright © 1996-2012 Alexander Bogomolny |
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