Date: 08/10/1999 at 00:32:43 From: Eric Evenson Subject: Napoleon's Triangle What is Napoleon's triangle?
Date: 08/10/1999 at 04:17:44 From: Doctor Floor Subject: Re: Napoleon's Triangle Dear Eric, Thanks for writing to Ask Dr. Math. Let ABC be a given triangle. Construct the point D such that ABD is an equilateral triangle, and D and C are on opposite sides of AB. In the same way, find point E such that BCE is equilateral, and point F such that CAF is equilateral. Let J be the center of ABD, K the center of BCE, and L the center of CAF. Then JKL is Napoleon's triangle: The interesting thing about Napoleon's triangle is that it is equilateral. To see this, find point O such that triangles DAO and ABC are (directly) congruent. In the same way, find point N such that BDN and ABC are directly similar. Construct Napoleon triangles DAO and BDN: Now consider triangle TEB in the picture. Note that BE = BC and BT = BN = AC. Note also that: <EBT = 360 - 3*60 - <ABC - <DBN. Since angle DBN is equal to angle BAC, we find that <EBT = 180 - <ABC - <BAC = <ACB. But that means that triangle TEB must be congruent to triangle ABC (SAS). We can conclude that, by the way it is constructed, GK = KL, and using the same reasoning show that GK = KL = LM = MI = IH = HG . Note that JL = JG by construction, GK = LK by , and of course JK = JK, so that triangles GJK and LJK are congruent (SSS). Thus <KJG = <LJK and we see that the six angles at J must all be congruent and are all 60 degrees. In particular, <LJK = 60 degrees. In the same way we can show that <JKL and <KLJ are 60 degrees, and thus triangle JKL is equilateral, as desired. Another interesting fact about Napoleon's triangle is described in the Dr. Math archives: The Napoleon Point and More http://mathforum.org/dr.math/problems/schultess9.4.98.html I hope this is what you were looking for. If you have any other math questions, send them to Dr. Math. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.