## 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 = 3GO (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 = 3GI. If we analogously define the Spieker centre S as the midpoint of NI, we also have SG = GI.