Nagel Point: What Is It About?
A Mathematical Droodle

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Nagel point

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Copyright © 1996-2018 Alexander Bogomolny

The applet may suggest the following statement:

Let E, D, F be the points of tangency of the excircles of ΔABC 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 ΔABC.

Nagel point, proof

The proof relies on Ceva's theorem.

First, let CC denote the excircle of ΔABC 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:


CT = CU.

For a similar reason




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 ΔABC 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.)

Related material

  • Incircles and Excircles in a Triangle
  • Nagel Point of the Medial Triangle
  • Homothety between In- and Excircles
  • Property of Points Where In- and Excircles Touch a Triangle
  • Feuerbach's Theorem: A Proof
  • Feuerbach's THeorem: What Is It?
  • |Activities| |Contact| |Front page| |Contents| |Geometry|

    Copyright © 1996-2018 Alexander Bogomolny