Three Congruent Circles by Reflection: What is this about?
A Mathematical Droodle

 

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Explanation

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Copyright © 1996-2017 Alexander Bogomolny

The applet attempts to illustrate a theorem by Quang Tuan Bui:

  In ΔABC, H is the orthocenter, Ab and Ac are reflections of A in the altitudes BH and CH. Ba, Bc, Ca, and Cb are defined similarly. Then the circumcircles of triangles ACbBc, BAcCa, and CBaAb are congruent.

 

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Proof

Let O be the circumcenter of ΔABC and X, Y, Z its reflections in AH, BH, CH, respectively. X is the circumcenter of triangle ABaCa, the reflection of triangle ABC in its A-altitude. It follows that H lies on the perpendicular bisector of OX so that

  HO = HX.

Similarly, HO = HY and HO = HZ. It follows that the four points O, X, Y, Z lie on a circle with center H.

Besides being equal, both triangles OBC and XCaBa are isosceles. Therefore, OB||XCa and also OB = XCa, making OXCaB a parallelogram. It follows that OX = BCa. Similarly, OX = CBa. Of course, analogues are also true for OY and OZ:

  OX = BCa = CBa
OY = CAb = ACb
OZ = ABc = BAc.

We thus obtain three pairs of equal triangles:

  ΔACbBc = ΔOYZ
ΔBAcCa = ΔOZX
ΔCBaAb = ΔOXY.

Note that the triangles in the right hand sides share the circumcircle. It follows that the circumcircles of the triangles in the left hand sides are congruent.

References

  1. Quang Tuan Bui, Two Triads of Congruent Circles from Reflections, Forum Geometricorum, Volume 8 (2008) 712.

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Copyright © 1996-2017 Alexander Bogomolny

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