### Hexagon Parallel to Orthic Triangle

On a side of an acute triangle, choose a point close to the foot of the altitude to that side. Starting with this point, draw a line parallel to one of the sides of the orthic triangle that meet at the foot of the altitude. Find the point of intersection of that line with another side of the triangle and, from there continue the process. The process generates a broken line. Prove that the sixth leg of this line terminates at the starting point. In other words, the process leads to a (closed) hexagon with sides parallel to the orthic triangle.

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Copyright © 1996-2018 Alexander Bogomolny

Let's exploit the mirror property of the orthic triangle. Reflect the triangle 5 times in successive sides as was done in the Schwarz's solution to the Fagnano's problem.

Due to the construction, images of the six legs of the broken line will form a straight line that connects a point (P) with its image under the five reflections. This exactly means that the hexagon is closed.

This hexagon has appeared in the context of the Tucker and Lemoine circles.

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Copyright © 1996-2018 Alexander Bogomolny

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