Three Congruent Circles by Reflection

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

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

In ΔABC, H is the orthocenter, A_b and A_c are reflections of A in the altitudes BH and CH. B_a, B_c, C_a, and C_b are defined similarly. Then the circumcircles of triangles AC_bB_c, BA_cC_a, and CB_aA_b 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 AB_aC_a, 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 XC_aB_a are isosceles. Therefore, OB||XC_a and also OB = XC_a, making OXC_aB a parallelogram. It follows that OX = BC_a. Similarly, OX = CB_a. Of course, analogues are also true for OY and OZ:

OX = BC_a = CB_a
OY = CA_b = AC_b
OZ = AB_c = BA_c.

We thus obtain three pairs of equal triangles:

ΔAC_bB_c = ΔOYZ
ΔBA_cC_a = ΔOZX
ΔCB_aA_b = 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

Quang Tuan Bui, Two Triads of Congruent Circles from Reflections, Forum Geometricorum, Volume 8 (2008) 7–12.

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