Simsons and 9-Point Circles in Cyclic Quadrilateral
What is this about?
A Mathematical Droodle


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(When the simsons are shown, you may want to verify their definition. Click on them in turn, or click away from any to remove additional information.)

Explanation

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Copyright © 1996-2017 Alexander Bogomolny

Explanation


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 http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

Any quadrilateral ABCD defines four triangles: BCD, CDA, DAB, ABC. In each of the triangles one may consider associate remarkable points and lines. The above applet is specifically concerned with their 9-point circles and also the Simson lines of each of the points A, B, C, D with respect to the triangle formed by the other three. The applet purports to illustrate the fact that, provided the quadrilateral ABCD is cyclic, the eight aforementioned elements all concur in a point.

First let's prove the concurrence of the four 9-point circles. As we know, the quadrilateral NANBNCND formed by the 9-point centers of the four triangles BCD, CDA, DAB, ABC is similar to ABCD with the factor of 1/2. So that the distance from the circumcenter of NANBNCND to its vertices is R/2, where R is the circumradius of ABCD. We also know that in any triangle the radius of its 9-point circle equals half that of its circumradius. It therefore follows that the circles of radii R/2 centered at the points NA, NB, NC, and ND meet at the circumcenter of the quadrilateral NANBNCND. In other words, the 9-point circles of the triangles BCD, CDA, DAB, ABC meet at the circumcenter of NANBNCND.

Now let's tackle the simsons. As we showed elsewhere, the circumcenter of NANBNCND coincides with H, the center of homothety that maps ABCD onto HAHBHCHD, the quadrilateral formed by the orthocenters of the triangles BCD, CDA, DAB, ABC. The two quadrilaterals ABCD and HAHBHCHD are equal so that H is the midpoint of each of the segments AHA, BHB, CHC, and DHD. By a well known property of triangles [Honsberger, p. 43], the simson of A with respect to ΔBCD passes through the midpoint of AHA, i.e., H. This of course applies to the other three simsons.

Note, that the result admits a natural generalization: if the quadrilateral is not cyclic, the simsons convert into circles, and the eight circles involved are still concurrent in a point.

References

  1. R. Honsberger, Episodes in Nineteenth and Twentieth Century Euclidean Geometry, MAA, 1995.

Related material
Read more...

Simson Line - the simson

  • Simson Line: Introduction
  • Simson Line
  • Three Concurrent Circles
  • 9-point Circle as a locus of concurrency
  • Miquel's Point
  • Circumcircle of Three Parabola Tangents
  • Angle Bisector in Parallelogram
  • Reflections of a Point on the Circumcircle
  • Simsons of Diametrically Opposite Points
  • Simson Line From Isogonal Perspective
  • Pentagon in a Semicircle
  • Simson Line in Disguise
  • Two Simsons in a Triangle
  • Carnot's Theorem
  • A Generalization of Simson Line

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