## Euler and Nagel lines for Cyclic and Circumscribed Quadrilaterals

1) **Euler line**: For a cyclic quadrilateral, a dilation of -1/3 with centre *G*, maps *ABCD* onto the centroid quadrilateral *A'B'C'D'*, and circumcentre *O* to *O'*. But a dilation with a scale factor of 3 from centre O, maps *A'B'C'D'* to *A"B"C"D"*, and *O"* to *H*. Therefore, if *H*, the circumcentre of *A"B"C"D"*, is defined as the 'orthocentre' of a cyclic quadrilateral, we have *HG* = 3*GO* (as well as *HP* = *PO*, where *P* is the nine-point (or Euler) centre).

2) **Nagel line**: For the circumscribed quadrilateral, a dilation of -1/3 with centre *G*, maps *ABCD* onto the centroid quadrilateral *A'B'C'D'*, and incentre *I* to *I'*, and because of the half-turn, *I*, *G* and *I'* are collinear. Next apply a dilation with a scale factor of 3 from centre *I*, to map *A'B'C'D'* to *A"B"C"D"*, and *I"* to the point *N*, which we now constructively define as the Nagel point of a circumscribed quadrilateral. Then from the applied transformations we have similarly to the cyclic case that *N*, *G* and *I* are collinear, and *NG* = 3*GI*. If we analogously define the Spieker centre *S* as the midpoint of *NI*, we also have *SG* = *GI*.

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Euler and Nagel lines for Cyclic and Circumscribed Quadrilaterals

Both results respectively generalize to cyclic and circumscribed polygons. For more information read my paper *Generalizing the Nagel line to Circumscribed Polygons by Analogy & Constructive Defining*.

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.