## Euler-Nagel line 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).

#### .sketch_canvas { border: medium solid lightgray; display: inline-block; } Euler-Nagel line 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 I, 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 for 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 about the above, read my 2008 Pythagoras paper Generalizing the Nagel line to Circumscribed Polygons by Analogy and 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.