Simson Line
A. Bogomolny

Given ΔABC and a point P, the feet of the perpendiculars from P to the sides of ΔABC form a triangle which is known as the pedal triangle of P. The definition apparently makes sense for any point P with no exception. However, according to a well known theorem proven in 1797 by W. Wallace, for P on the circumcircle of ΔABC, the feet of the perpendiculars are colinear. The line is known as the Simson line of P, or just simson. This definition is somewhat strange, for nowhere in the works of Robert Simson (1687-1768) any reference was found to the line that bears his name. On the other hand, he is said [F. G.-M., p. 496] to have discovered the theorem that bears the name of Stewart. He's also proved [Coxeter, p. 41] the identity Fn+1Fn+1 - FnFn+2 = (-1)n between the Fibonacci numbers that explains the Fibonacci Bamboozlement puzzle.

28 November 2015, Created with GeoGebra

Back to the Simson line, do the points on the circumcircle have pedal triangles? Instead of making an exceptional case out of the points on the circumcircle, mathematicians found it convenient to talk of degenerate triangles, triangles with colinear vertices. By that definition, every point has an associated pedal triangle. For the points on the circumcircle, the pedal triangle degenerates into a line. That's all. Sometimes, Wallace's theorem is not discussed with pedal triangles in the background. On such occasions, the whole point of degenerate triangles becomes moot.

But having degenerate pedal triangles sets points on the circumcircle aside from other points in the plane. There is another property peculiar to these points.

28 November 2015, Created with GeoGebra

Let P lie on the circumcircle of ΔABC. Reflect AP in the angle bisector of angle A and extend the line in both directions. Similarly, find the reflection of BP in the bisector of B and that of CP in the bisector of C. It's a curious fact that the three lines thus obtained are parallel.

And what about points P not on the circumcircle? For such points, the three reflections are concurrent. The point of concurrency Q is defined as the isogonal conjugate of P. By symmetry, if Q is the isogonal conjugate of P, then P is the isogonal conjugate of Q. Whose isogonal conjugates are points on the circumcircle? By definition, points on the circumcircle are conjugate to points at infinity, one point at infinity per direction. All together, the points at infinity form a line, the line at infinity. There is no possible ambiguity. A bundle of all possible directions forms a line (at infinity), with the lines parallel to one of the directions intersecting at a single point (at infinity). In trilinear coordinates, the line at infinity looks like any other line, see [Kimberling, p. 28].

28 November 2015, Created with GeoGebra

Let's regress a little. Points on the circumcircle have associated Simson lines. The feet of the perpendiculars from points outside the circumcircle to the sides of the triangle are not colinear. They do not lie on a straight line. But, as for other triples of non colinear points, there exists a unique circle to which all three belong. As it comes out, those circles for P and its isogonal conjugate Q coincide. Now six points - the feet of the perpendiculars from two isogonal conjugates to the sides of the triangle - are concyclic, all six belong to the same circle. By definition, the circumcircle of a pedal triangle is called a pedal circle. The above mentioned theorem can be seen as asserting that isogonally conjugate points share the same pedal circle.

When P belongs to the circumcircle, its isogonal conjugate Q goes to infinity. By definition, the circle at hand acquires infinite radius. It becomes a straight line - the simson of point P.

Simson lines have many interesting properties and are also useful in establishing other facts. For example, existence of Simson lines together with the Pivot theorem could be used to demonstrate the existence of Miquel's point for a 4-line. In passing, the Pivot theorem was discovered by A. Miquel in 1838. It was so named by H. G. Forder [Geometry Revisited, p. 62] If the point of intersection (the Pivot point) is joined to the points on the sidelines of the triangle, then these lines are equally inclined to the sides. And this property fully characterizes the pivot point. Therefore, if the three lines are rotated around the pivot point through the same angle, the newly obtained circles of the Pivot theorem still all pass through the same point.

28 November 2015, Created with GeoGebra

The Simson line also have an interesting generalization. To obtain the Simson line of point P on the circumcircle, three lines are drawn from P that make angles of 90° with the sides of a given triangle. A theorem by Lasare Carnot (1753-1823) states that, if the lines are drawn to make any, not necessarily a 90°, angle with the sides (orientation matters), the points where they cross the sides will be colinear. The common line may be called the oblique Simson line.

28 November 2015, Created with GeoGebra

What we observe here again relates to the Pivot theorem. Rotate the three lines joining the pivot to the points on the sidelines through the same angle. Watch the points of intersection of the three lines with the sidelines of the triangle. If in one configuration the three points were colinear, they will remain colinear after any rotation around the pivot point.

Wallace's and Carnot's theorems provide a nice example of a mathematical fact and its generalization where proving the generalization is as easy (or as difficult) as proving the particular case. However, simple as the generalization may appear, it truly opens new horizons for further investigation. For example, for no point may the pedal triangle coincide with the given one. However, once we allow for oblique pedal triangles suggested by Carnot's theorem the situation changes. There exist two points - Brocard points - (one per orientation) whose oblique pedal triangles coincide for a certain angle - the Brocard angle - with the given triangle. The Brocard angle is the same for both orientations.

Last but not least, the Simson line relates to other remarkable elements in a triangle. As an example, consider the Simson line of a point P and the diametrically opposite to P point Q.

28 November 2015, Created with GeoGebra

The Simson lines of P and Q are orthogonal. Remarkably, the point of intersection of the two lines lies on the 9-point circle of the given triangle. The oblique simsons of P and Q (with the same inclination) are also orthogonal. As P moves over the circumcircle, the point of intersection of the two lines traces a circle, the 9-point circle in the case of the (regular) Simson lines.

28 November 2015, Created with GeoGebra


  1. H.S.M.Coxeter, S.L.Greitzer, Geometry Revisited, MAA, 1967
  2. F. G.-M., Exercise de Géométrie, Éditions Jacques Gabay, 1991
  3. C. Kimberling, Triangle Centers and Central Triangles, Utilitas Mathematica Publishing Inc., 1998
  1. Necessary and Sufficient
    (An attempt to draw conclusions from a remarkable experiment of more than half a century ago and some more recent ideas.)

  2. The Nature of Proof
    (A report on the above experiment.)

  3. Simson Line
    (A sequence of nice geometric facts with the word define emphasized. Just imagine what would happen if we did not agree on the definitions or did not use them altogether.)

Related material

Simson Line - the simson

  • Simson Line: Introduction
  • Three Concurrent Circles
  • 9-point Circle as a locus of concurrency
  • Miquel's Point
  • Circumcircle of Three Parabola Tangents
  • Angle Bisector in Parallelogram
  • Simsons and 9-Point Circles in Cyclic Quadrilateral
  • 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

    |Contact| |Front page| |Contents| |Geometry| |CTK|

    Copyright © 1996-2018 Alexander Bogomolny