# Two Simsons in a Triangle

What Is This About?

A Mathematical Droodle

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Copyright © 1996-2018 Alexander Bogomolny

The applet attempts to illustrate a generalization of a property of the simsons corresponding to two diametrically opposite points on a circumcircle of a triangle:

For two diametrically opposite points on a circumcircle of a triangle the two simsons are perpendicular. The simsons of any two points P and Q on a circumcircle of ΔABC meet at an angle equal to one of the angles formed by P and Q and one of the vertices of the triangle. The proof below is by Hubert Shutrick.

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More accurately, let sP and sQ be the simson lines of points P and Q and M the intersections of the two. P lies on an arc of the circumcircle opposite one of the vertices A, B, or C which we'll denote vP and similarly introduce vQ. If

Define also, S to be the intersection of sP and vQ and T the intersection of sQ and vP. What Professor Shutrick has found is that the quadrilateral MSVT is cyclic. The proof depends on a particular configuration of the five points A, B, C, P, Q. Below we shall consider just one case:

In this case, V = A, S is the intersection of sP and AQ and T is the intersection of sQ and AP. Other notations have been borrowed from the case of diamterically opposite P and Q.

Since quadrilaterals AP_{b}P_{c}P and AQ_{c}Q_{b}Q are cyclic,

∠PAP_{c} = ∠PP_{b}P_{c} and

∠QAQ_{b} = ∠QQ_{c}Q_{b}.

By the construction, ∠AP_{b}P = ∠AQ_{c}Q = 90° implying that in quadrilateral AP_{b}MQ_{c},

∠Q_{c}MP_{b} | = 360° - ∠BAC - ∠AP_{b}M - ∠AQ_{c} | |

= 360° - ∠BAC - (90° + ∠PP_{b}P_{c}) - (90° + ∠QQ_{c}Q_{b}) | ||

= 180° - ∠BAC - ∠PAP_{c} - ∠QAQ_{b}) | ||

= 180° - ∠PAQ. |

From here, ∠SAT + ∠SMT = 180° and the quadrilateral ASMT is indeed cyclic.

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

Copyright © 1996-2018 Alexander Bogomolny

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