   too complicated to be written here. Click on the link to download a text file.  X(111), X(671), X(2482), X(6593) midpoints A', B', C' of ABC midpoint E = X(6593) of X(6), X(110) other points below    K565 is a psK parent of the Thomson cubic K002. As such, it contains the midpoints of ABC and the tangents at A, B, C are the symmedians. It meets the circumcircle again at X(111) and two other points Q2, Q3 on the tangent at X(2482) to the Steiner inscribed ellipse. It meets the Steiner inscribed ellipse at A', B', C', X(2482) and two other points R2, R3 on the polar line of X(111) in this ellipse. It meets the Steiner ellipse at A, B, C, X(671) and two other points S2, S3 on the line passing through E and X(690). The tangents at X(111), Q2, Q3 are concurrent. *** Note about X(2482) X(2482) is the antipode of X(115) on the Steiner inscribed ellipse, the complement of X(671), the midpoint of X(2)-X(99). It is the only point (apart A', B', C') on the Steiner inscribed ellipse such that the tangent (to this ellipse) meets the circumcircle at two points (here denoted by Q2, Q3) whose midpoint is the point of tangency of this latter tangent. X(2482) is also the foot of the fourth normal drawn through O to the Steiner inscribed ellipse.  