| Manifesto | Bookstore | Contents | Amazon store | Term index | What changed? | Contact | Recommend |
Two Congruent Circles by Reflection: What is this about?
|
| Buy this applet What if applet does not run? |
The applet attempts to illustrate a problem by Quang Tuan Bui:
| I is the incenter and X an arbitrary point in the plane of ΔABC. Bi and Bx are reflections of B in AI and AX, and similarly Ci and Cx. Show that the circumcircles CBiBx and BCiCx are congruent. |
| Buy this applet What if applet does not run? |
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 Ci is obtained from Cx by a rotation around A which also maps Bi onto Bx.
It follows that triangles CiACx and BiABx are both isosceles with the same apex angle which makes their base angle also equal. In particular,
 ACiCx = ABiBx.
|
Now, let Bj be the reflection of Bi in AX. We have
|
implying that the quadrilateral BCiCxBj is cyclic. The circle circumscribed about that quadrilateral is a circumcircle of two triangles, BCiCx and BCxBj. But ΔBCxBj is the reflection of ΔBxCBi in AX so that their circumcircles are congruent. This shows that the circumcircles BCiCx and CBiBx are also congruent.
Naturally, the two circles are the reflections of each other in AX.
| 35688907 |  |
