Three Congruent Circles by Reflection: What is this about?
A Mathematical Droodle
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Copyright © 1996-2018 Alexander BogomolnyThe 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_{b}B_{c}, BA_{c}C_{a}, and CB_{a}A_{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_{a}C_{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_{a}B_{a} are isosceles. Therefore, OB||XC_{a} and also OB = XC_{a}, making OXC_{a}B a parallelogram. It follows that
OX = BC_{a} = CB_{a} OY = CA_{b} = AC_{b} OZ = AB_{c} = BA_{c}. |
We thus obtain three pairs of equal triangles:
ΔAC_{b}B_{c} = ΔOYZ ΔBA_{c}C_{a} = ΔOZX ΔCB_{a}A_{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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Copyright © 1996-2018 Alexander Bogomolny64639424 |