## 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).

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