Simson line

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The Simson line of the triangle ABC.

In geometry, given a triangle and a point on its circumcircle, the intersections formed when lines are constructed from the point perpendicular to each of the triangle's sides are collinear. The line through these points is the Simson line, named for Robert Simson.^[1]

The converse is also true; if the feet of the perpendiculars dropped from a point to the sides of the triangle are collinear, then the point is on the circumcircle. The Simson line of a point is just the pedal triangle of it; the case when that pedal triangle degenerates to a line.

1 Properties
2 Proof of existence
3 See also
4 References
5 External links

[edit] Properties

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 $P'$ are points on the circumcircle, then the angle between the Simson lines of $P$ and $P'$ is half the angle of the arc $P P'$ . In particular, if the points are diametrically opposite, their Simson lines are perpendicular and in this case the intersection of the lines is on the nine-point circle.

Let $H$ denote the orthocenter of the triangle $A B C$ , then the Simson line of $P$ bisects the segment $P H$ 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 doesn't depend of $P$ .

[edit] Proof of existence

The method of proof is to show that $\angle MNP + \angle PNL = 180^\circ$ . $P B C A$ is a cyclic quadrilateral, so $\angle PAM + \angle CBP = \angle PAC + \angle CBP = 180^\circ$ . $P N M A$ is a cyclic quadrilateral (Thales' theorem), so $\angle PAM + \angle MNP = 180^\circ$ . Hence $\angle MNP = \angle CBP$ . Now PLBN is cyclic, so $\angle PNL = \angle PBL = 180^\circ - \angle CBP$ . Therefore $\angle MNP + \angle PNL = \angle CBP + (180^\circ - \angle CBP) = 180^\circ$ .