Reflections of a Line Through the Orthocenter
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A Mathematical Droodle


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Explanation

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Copyright © 1996-2018 Alexander Bogomolny

The following result is due to L. Carnot (1801):

Reflect a line passing through the orthocenter H of ΔABC in the three sidelines. The three reflections concur on the circumcircle of the triangle.

The problem quite easily solved by "angle chasing" - counting angles which, due to the layout of the problem, come in just several related sizes. However, configurations differ in details and several cases ought to be considered. That said, the idea of the proof can be introduced with a particular case:

Assume a line through the orthocenter H of ∠ABC crosses AC at Tb and BC at Ta. Denote angle CTbH as β and angle BTaH as α. Then

(1) α + β + ∠ACB = 180°.

As we know, reflection Ka of the orthocenter H in BC lies on the circumcircle, as does its reflection Kb in AC. Next observe that the lines TaKa and TbKb are exactly the reflections of the given line in BC and AC, respectively. Assume the two meet in Q and let's show that Q lies on the circumcircle of the given triangle.

In ∠QTaT angle Ta is obviously 180° - 2α, while angle Tb measures 180° - 2β. Thus

∠TaQTb = 180° - (180° - 2α) - (180° - 2β)
  = 2(α + β) - 180°
  = 2(180° - ∠ACB) - 180°
  = 2(90° - ∠ACB),

where we applied (1) on the last step. But this is exactly half the angular measure of arc KaCKb that subtends the angle KaQKb. The latter is therefore inscribed into the circumcircle, so that indeed this is where Q lies and TaKa and TbKb meet.

There are two more pairs of reflections to consider, but since each reflection participates in two pairs but intersects the circle only once (not counting the points K), all three pairs meet in the same point which is bound to be Q.

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Copyright © 1996-2018 Alexander Bogomolny
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