Three Congruent Circles by Reflection: What is this about?
A Mathematical Droodle
| |
|
Explanation
Copyright © 1996-2008 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.
|
| |
|
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
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
- Quang Tuan Bui, Two Triads of Congruent Circles from Reflections, Forum Geometricorum, Volume 8 (2008) 7–12.
Copyright © 1996-2008 Alexander Bogomolny
|