Two Congruent Circles by Reflection: What is this about?
A Mathematical Droodle
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Copyright © 19962018 Alexander BogomolnyThe applet attempts to illustrate a problem by Quang Tuan Bui:
I is the incenter and X an arbitrary point in the plane of ΔABC. B_{i} and B_{x} are reflections of B in AI and AX, and similarly C_{i} and C_{x}. Show that the circumcircles CB_{i}B_{x} and BC_{i}C_{x} are congruent.
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Proof
The product of the reflections in two intersecting axes is a rotation through twice the angle between them, with the center at the point of intersection. So that C_{i} is obtained from C_{x} by a rotation around A which also maps B_{i} onto B_{x}.
It follows that triangles C_{i}AC_{x} and B_{i}AB_{x} are both isosceles with the same apex angle which makes their base angle also equal. In particular,
∠AC_{i}C_{x} = ∠AB_{i}B_{x}. 
Now, let B_{j} be the reflection of B_{i} in AX. We have

implying that the quadrilateral BC_{i}C_{x}B_{j} is cyclic. The circle circumscribed about that quadrilateral is a circumcircle of two triangles, BC_{i}C_{x} and BC_{x}B_{j}. But ΔBC_{x}B_{j} is the reflection of ΔB_{x}CB_{i} in AX so that their circumcircles are congruent. This shows that the circumcircles BC_{i}C_{x} and CB_{i}B_{x} are also congruent.
Naturally, the two circles are the reflections of each other in AX.
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Copyright © 19962018 Alexander Bogomolny71411378