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.
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.
Michael de Villiers, 6 April 2010.