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Nagel Point: What Is It About?
A Mathematical Droodle


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


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Explanation

Copyright © 1996-2008 Alexander Bogomolny

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The applet may suggest the following statement:

 

Let E, D, F be the points of tangency of the excircles of ABC and its sides, as in the applet below. Prove that AD, BE, CF are concurrent. The point of concurrency is known as the Nagel point of ABC.



This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


Buy this applet

The proof relies on Ceva's theorem.

First, let CC denote the excircle of ABC opposite the vertex C. CC is tangent to the side AB and the side lines AC and BC in points F, T and U. Since, CT and CU are the two tangents from C to CC, they are equal in length:

(1) CT = CU.

For a similar reason

(2)AT = AF, and
(3)BU = BF.

(1)-(3) immediately imply that F is a perimeter splitter in the sense that

  AC + AF = BC + BF.

If a, b, c are the side lengths of the ABC and p its semiperimeter in a standard way, then this fact could be rewritten as

  b + AF = a + BF = p,

so that

  AF = p - b
BF = p - a.

Similarly, we obtain additional four identities:

  AE = p - c
CE = p - a
CD = p - b
BD = p - c.

Ceva's identity is then verified directly:

  AE/CE·CD/BD·BF/AF = (p-c)/(p-a)·(p-b)/(p-c)·(p-a)/(p-b) = 1.

(A curious fact is that these same points D, E, F arise in a related, yet quite a different construction.)

Copyright © 1996-2008 Alexander Bogomolny

28696803Page copy protected against web site content infringement by Copyscape


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