Date: 04/19/99 at 21:28:21 From: Frank Furman Subject: Simpson's Line How do you get Simpson's line (geometry construction)?
Date: 04/20/99 at 08:51:03 From: Doctor Floor Subject: Re: Simpson's Line Hi, Frank, Thanks for your question. Let ABC be a triangle with a point D on its circumcircle. When we construct the feet of D on the sides of ABC (extended if necessary), these three feet are on one line, the Simson line (note the spelling) or Wallace line. You can see this in the following picture: Basic constructions for compass and straightedge that might be helpful can be found in the Dr. Math archives: Geometry Constructions with Compass and Straightedge http://mathforum.org/dr.math/problems/zaidi11.13.98.html Constructing the Orthocenter http://mathforum.org/dr.math/problems/justin1.27.99.html Also, to construct the circumcircle of a triangle it is good to know that the center of this circle (the circumcenter) is the intersection of the three perpendicular bisectors of the sides of triangle ABC. How do we prove that the feet of the altitudes of D are on one line? First we note that angle DEB = angle DGB = 90 degrees, so EDBG is a cyclic quadrilateral. In the same way we find that DFBC is a cyclic quadrilateral. From this we can find a couple of angle equalities: So we have: angle DGE = angle DBE = angle DBA = - angle ABD (these are directed angles!) = - angle ACD = angle DCA = angle DGF and we can conclude that angles DGF and DGE are equal, and so E, F, and G must be on one line. If you have a math question again, please send it to Ask Dr. Math. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
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