too complicated to be written here. Click on the link to download a text file. X(4), X(9), X(40), X(55), X(57), X(144) Foci of the Mandart ellipse (see Generalized Mandart Conics in the Downloads page and also K040) A', B', C' : vertices of the extouch triangle, cevian triangle of X(8)
 K710 is a circum-cubic that passes through the foci of the Mandart (inscribed) ellipse which are also the foci of the Mandart (circum) ellipse, the circum-conic with center X(9) and perspector X(1). These two conics (together with their extraversions) are the only conics having this property, see a proof below. K710 meets : • the line at infinity at the same points as pK(X6, X7) or pK(X1, X3434), • the Mandart (circum) ellipse at the same points as pK(X1, X63), • the circumcircle at the same points as pK(X6, X44). See K040, K351 = pK(X6, X9), K352 also passing through these foci. K710 is also spK(X7, X9) in CL055. See the related K716. *** Proof : let I(P) be the in-conic with perspector P and center Q = ctP. Let C(cP) be the circum-conic with perspector cP and same center Q as I(P). Following Poncelet, these two conics have the same foci if and only if the two pairs of tangents drawn from any point S to both conics have the same bisectors. Taking S = A, the tangents to I(P) are the bisectors at A in ABC and the tangents to C(cP) coincide hence the tangent at A to C(cP) must be one of the bisectors at A in ABC. Similarly with S = B and S = C we obtain that cP must be an in/excenter of ABC. It follows that P, tP and ctP must be X(8), X(7) and X(9) respectively or one of the corresponding extraversions. Finally, I(P) must be the Mandart ellipse or one of its extraversions.