# 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.

28 November 2015, Created with GeoGebra

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 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.

28 November 2015, Created with GeoGebra

Let the feet of the perpendiculars from P to BC, CA and AB be A_{P}, B_{P}, C_{P}, and similarly define A_{Q}, B_{Q}, C_{Q}. The above states that the circumcircles of triangles A_{P}B_{P}C_{P} and A_{Q}B_{Q}C_{Q} coincide. In other words, we arrive at the following

### Theorem

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

### Proof

The quadrilateral PB_{P}AC_{P} is cyclic, for two opposite angle at B_{P} and C_{P} both measuring 90° add up to 180°. Therefore two angles PB_{P}C_{P} and PAC_{P} that subtend the same chord PC_{P} are equal. Similarly, _{Q} = ∠QC_{Q}B_{Q}._{P} = ∠QAB_{Q}. Therefore, _{P}C_{P} = _{Q}B_{Q}._{P}A and QB_{C}A we obtain _{P}B_{P}B_{Q} = _{Q}C_{Q}C_{P}._{Q}C_{P}, which proves that the four points B_{P}, B_{Q}, C_{P}, C_{Q} are concyclic - they lie on the same circle. The center of the circle is necessarily the point of intersection of the perpendicular bisectors to B_{P}B_{Q} and C_{P}C_{Q}, which is the midpoint of the segment PQ.

The above argument is completely symmetric with respect to the sides of ΔABC. So it follows that all six points A_{P}, A_{Q}, B_{P}, B_{Q}, C_{P} and C_{Q} lie on a circle centered at the midpoint of PQ.

As we know, if P lies on the circumcircle of ΔABC, 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 ΔABC, 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 ΔABC.

### References

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

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny

61236457 |