## Euler-Nagel analogy

1) The **circumcenter** (O), centroid (G) & **orthocenter** (H) of any triangle ABC are collinear (**Euler** line), GH = 2GO and the midpoint of OH is the center of the **nine-point** circle (P).

2) The *incenter* (I), centroid (G) & *Nagel* point (N) of any triangle are collinear (*Nagel* line), GN = 2 GI and the midpoint of IN is the center of the *Spieker* circle (S).
####

Euler-Nagel analogy

a) Note that for the Euler line, a halfturn with centre *G* and a scale factor of 1/2, maps *ABC* onto the median triangle *A'B'C'*, and circumcentre *O* to *P*. But a dilation with a scale factor of 2 from centre *O*, maps *A'B'C'* to *A"B"C"*, and *P* to *H*. Therefore, *H* (the orthocentre of *ABC*) is the circumcentre of *A"B"C"*.

b) Similarly, for the Nagel line, a halfturn with centre *G* and a scale factor of 1/2, maps *ABC* onto the median triangle *A'B'C'*, and incentre *I* to *S*. But a dilation with a scale factor of 2 from centre *O*, maps *A'B'C'* to *A"B"C"*, and *S* to *N*. Therefore, *N* (the Nagel point of *ABC*) is the incentre of *A"B"C"*.

Since the Euler line generalizes to cyclic polygons, a Nagel line to circumscribed polygons can now be generalized by the above analogy as shown by the dynamic sketch at *Nagel line for circumscribed quadrilateral*

For more information read my paper *Generalizing the Nagel line to Circumscribed Polygons*.

Related link 1: For a dynamic sketch of the centroid *G* of a quadrilateral go to *Point Mass Centroid of a quadrilateral*.

Related link 2: For a dynamic sketch of the nine-point (or Euler) centre *P* of a cyclic quadrilateral go to *Nine-point (Euler) centre of a cyclic quadrilateral*.

This page uses **JavaSketchpad**, a World-Wide-Web component of *The Geometer's Sketchpad.* Copyright © 1990-2008 by KCP Technologies, Inc. Licensed only for non-commercial use.

Michael de Villiers, 6 April 2010.