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Three Congruent Circles by Reflection: What is this about?
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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
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OX = BCa = CBa OY = CAb = ACb OZ = ABc = BAc. |
We thus obtain three pairs of equal triangles:
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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
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