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Simson Line From Isogonal Perspective

If a point is selected on the circumcircle of a triangle and the perpendiculars are dropped from that point to the sides of the triangle, the feet of the perpendiculars are colinear. The line to which the three belong is known as the Simson line (of the selected point with respect to the triangle.)

The applet below illustrates a property of isogonal conjugates that generalizes the Simson line.


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.


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Explanation

Copyright © 1996-2008 Alexander Bogomolny

 

 

 

 

 

 

 

 

 

 

 

 

 

The six feet of the perpendiculars to the sides of triangle ABC dropped from a point P and its isogonal conjugate Q are concyclic, i.e. all six lie on the same circle.


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.


Buy this applet

Let the feet of the perpendiculars from P to BC, CA and AB be AP, BP, CP, and similarly define AQ, BQ, CQ. The above states that the circumcircles of triangles APBPCP and AQBQCQ coincide. In other words, we arrive at the following

Theorem

The pedal circles of a point and its isogonal conjugate coincide.

Proof

The quadrilateral PBPACP is cyclic, for two opposite angle at BP and CP both measuring 90o add up to 180o. Therefore two angles PBPCP and PACP that subtend the same chord PCP are equal. Similarly, QABQ = QCQBQ. Since P and Q are isogonal conjugate, we also have PACP = QABQ. Therefore, PBPCP = QCQBQ. Subtracting these from the equal angles PBPA and QBCA we obtain CPBPBQ = BQCQCP. The latter angles subtend the same segment BQCP, which proves that the four points BP, BQ, CP, CQ are concyclic - they lie on the same circle. The center of the circle is necessarily the point of intersection of the perpendicular bisectors to BPBQ and CPCQ, which is the midpoint of the segment PQ.

The above argument is completely symmetric with respect to the sides of ABC. So it follows that all six points AP, AQ, BP, BQ, CP and CQ lie on a circle centered at the midpoint of PQ.

As we know, if P lies on the circumcircle of ABC, Q is a point at infinity. In that case, the radius of the above circle is infinite, and the circle appears to be a line. Such that, if P lies on the circumcircle of ABC, the feet of the three perpendiculars from P to the sides of the triangle lie on that straight line. The line is known as the Simson line of P with respect to ABC.

References

  1. R. Honsberger, Episodes in Nineteenth and Twentieth Century Euclidean Geometry, MAA, 1995, pp. 67-68
  2. V. V. Prasolov, Essays On Numbers And Figures, AMS, 2000, pp. 65-69

Copyright © 1996-2008 Alexander Bogomolny

28761889Page copy protected against web site content infringement by Copyscape


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