The Simson Line

The sketch below is constructed as follows:  For a triangle DABC, we construct the circumcircle and choose a point P on this circle.  The point D is the point on the (extended) side BC such that PD is perpendicular to BC.  The points E and F are similarly defined.  (Don't take my word for this -- check that these are infact perpendicular.)  A priori, there is no reason to expect that the three points D, E and F are colinear, but the sketch infact suggests that they are.

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Move the point P around the circle.  Note that the three points remain colinear (and are still the bases of perpendiculars through P).  The triangle, too, can be changed -- move any one of the vertices around.  This sketch is a demonstration of the following theorem:

Theorem:  Let P be a point on the circumcircle of an arbitrary triangle DABC.  Let D, E and F be the bases of perpendiculars through P to the (extended) sides BC, AC and AB respectively.  Then D, E and F are colinear.  This line is called the Simson line.

The java applet above was produced by JavaSketchpad.  The following is the maker's beta:

This is a prototype of JavaSketchpad, a World-Wide-Web component of The Geometer's Sketchpad. Copyright ©1990-1998 by Key Curriculum Press, Inc. All rights reserved. Portions of this work were funded by the National Science Foundation (awards DMI 9561674 & 9623018).

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This page was created November 26th, 1998.