Placing the triangle in the complex plane, let the triangle ABC with unit circumcircle have vertices whose locations have complex coordinates a, b, c, and let P with complex coordinates p be a point on the circumcircle. The Simson line is the set of points z satisfying[4]^{:Proposition 4}

${\displaystyle 2abc{\bar {z}}-2pz+p^{2}+(a+b+c)p-(bc+ca+ab)-{\frac {abc}{p}}=0,}$

where an overbar indicates complex conjugation.

Simson lines (in red) are tangents to the Steiner deltoid (in blue).

• The Simson line of a vertex of the triangle is the altitude of the triangle dropped from that vertex, and the Simson line of the point diametrically opposite to the vertex is the side of the triangle opposite to that vertex.
• If P and Q are points on the circumcircle, then the angle between the Simson lines of P and Q is half the angle of the arc PQ. In particular, if the points are diametrically opposite, their Simson lines are perpendicular and in this case the intersection of the lines lies on the nine-point circle.
• Letting H denote the orthocenter of the triangle ABC, the Simson line of P bisects the segment PH in a point that lies on the nine-point circle.
• Given two triangles with the same circumcircle, the angle between the Simson lines of a point P on the circumcircle for both triangles does not depend of P.
• The set of all Simson lines, when drawn, form an envelope in the shape of a deltoid known as the Steiner deltoid of the reference triangle.
• The construction of the Simson line that coincides with a side of the reference triangle (see first property above) yields a nontrivial point on this side line. This point is the reflection of the foot of the altitude (dropped onto the side line) about the midpoint of the side line being constructed. Furthermore, this point is a tangent point between the side of the reference triangle and its Steiner deltoid.
• A quadrilateral that is not a parallelogram has one and only one pedal point, called the Simson point, with respect to which the feet on the quadrilateral are collinear.[5] The Simson point of a trapezoid is the point of intersection of the two nonparallel sides.[6]^{:p. 186}
• No convex polygon with at least 5 sides has a Simson line.[7]

The method of proof is to show that ${\displaystyle \angle NMP+\angle PML=180^{\circ }}$. ${\displaystyle PCAB}$ is a cyclic quadrilateral, so ${\displaystyle \angle PBA+\angle ACP=\angle PBN+\angle ACP=180^{\circ }}$. ${\displaystyle PMNB}$ is a cyclic quadrilateral (Thales' theorem), so ${\displaystyle \angle PBN+\angle NMP=180^{\circ }}$. Hence ${\displaystyle \angle NMP=\angle ACP}$. Now ${\displaystyle PLCM}$ is cyclic, so ${\displaystyle \angle PML=\angle PCL=180^{\circ }-\angle ACP}$. Therefore ${\displaystyle \angle NMP+\angle PML=\angle ACP+(180^{\circ }-\angle ACP)=180^{\circ }}$.

Alternate proof

green line is the Simpson's line, the blue ones are the perpendiculars dropped.

Whatever the point Z in the adjacent figure is, a + c is 90. Also, whatever the point Z is, c and b are going to be equal. Therefore, we have the following:

a + c = 90

∴ a + b = 90 …(c and b are equal) (1)

Now, consider the measure of angle: a + 90 + b.

If we show that this angle is 180, then the Simpson’s theorem is proved.

From (1) we have, a + 90 + b = 180

Q.E.D.

• Pedal triangle
• Robert Simson

Wikimedia Commons has media related to Simson line.

• Simson Line at cut-the-knot.org
• F. M. Jackson and Weisstein, Eric W. "Simson Line". MathWorld.
• A generalization of Neuberg's theorem and the Simson-Wallace line at Dynamic Geometry Sketches, an interactive dynamic geometry sketch.