Miquel's Theorem for Circles

Let there be two triples of points A0, A1, A2 and B0, B1, B2. The applet below illustrates the following statement: if circles A0B1B2, B0A1B2, B0B1A2 concur in a point V different from any of the given six, then the same holds for the circles B0A1A2, A0B1A2, A0A1B2.


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

Proof

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Copyright © 1996-2017 Alexander Bogomolny

Miquel's theorem for circles

The statement is a consequence of Miquel's theorem and basic properties of inversion. Indeed, suppose circles A0B1B2, B0A1B2, B0B1A2 meet in point V. An inversion with center V converts the three circles into three straight lines that form a triangle B'0B'1B'2 with images B' of points B and points A' - the images of points A - on the sides of this triangle. According to Miquel's theorem, the circumcircles of triangles B'0A'1A'2, A'0B'1A'2, A'0A'1B'2 meet in a point, the Miquel point of the configuration. The latter is the inversive image of the common point of circles B0A1A2, A0B1A2, A0A1B2.

Inversion - Introduction

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Copyright © 1996-2017 Alexander Bogomolny

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