   too complicated to be written here. Click on the link to download a text file.  X(4), X(12523), X(14529) A'B'C' : circumcevian triangle of X(1) HaHbHc : cevian triangle of X(4) in A'B'C' Z = X(14529) = H* isogonal conjugate of H wrt A'B'C' ZaZbZc pedal triangle of Z in ABC infinite points of the altitudes X = X(12523) = incenter of A'B'C' Xa, Xb, Xc excenters of A'B'C' (not represented)    Let A'B'C' be the circumcevian triangle of the incenter X(1) of ABC also called 2nd circumperp triangle, TCCT 6.22. There is a unique isogonal pK with respect to A'B'C' which is also a circum-cubic of ABC (although it is not a pK) and this cubic is K838. The union of the internal bisectors of ABC can be considered as pK(X6, X1) hence K838 is a special case of the cubics in Table 58. The pivot of K838 is H and its isopivot Z is the isogonal conjugate of H with respect to A'B'C'. Z = X(14529) lies on the lines {X1,X1437}, {X6,X2333}, {X31,X56}, {X40,X692}, {X47,X859}, {X65,X184}, etc. Z is also the midpoint of X(221), X(3556) and X(3157), X(9798). Recall that Z is the common tangential of H, A', B', C' in K838. K838 has three real asymptotes perpendicular to the sidelines of ABC hence parallel to those of the Darboux cubic K004. These two cubics have six other finite points namely A, B, C, H and S1, S2. These two latter (not always real) points lie on the rectangular circum-hyperbola (H) passing through X(103) and on the line (L) passing through X(1498) on K004 and Z on K838. K838 meets the sidelines of : • ABC at Za, Zb, Zc which are the vertices of the pedal triangle of Z in ABC, • A'B'C' at Ha, Hb, Hc which are obviously the vertices of the cevian triangle of H in A'B'C'. Note that these six latter points lie on a same conic (C) with center X(1125). K838 must contain the incenter X and the excenters of A'B'C' with tangents passing through the pivot H. This point X is now X(12523) in ETC with first barycentric coordinate : a [a^2 Sqrt(b c sb sc) - b (a+c) Sqrt(a c sa sc) - c (a+b) Sqrt(a b sa sb)] X is the midpoint of X(1), X(164). The orthic line of K838 is the Euler line (of ABC). Hence, the polar conic of H is the rectangular diagonal (in A'B'C') hyperbola passing through H and the the in/excenters of A'B'C'.   