1. The triangle NaNbNc bounded by the lines BaCa, CbAb, AcBc is homothetic to triangle ABC. The center of homothety is the symmedian point K of ABC and we have KNa:KA = KNb:KB = KNc:KC = (a^2 + b^2 + c^2 + S): S, being S twice the area of ABC.
2. If Ma, Mb, Mc are the midpoints of line segments ANa, BNb, CNc the medians of triangle MaMbMc are the perpendicular bisectors of line segments AbAc, BcBa and CaCb, respectively. Hence these perpendicular bisectors concur at centroid M of MaMbMc.
3. The triangle MaMbMc is thus orthological to triangle LaLbLc bounded by the lines AbAc, BcBa and CaCb, and the centroid M of MaMbMc is the center of orthology, hence the triangle LaLbLc is orthological to triangle MaMbMc being the centroid L of LaLbLc the center of orthology, that is, the perpendiculars to lines MbMc, McMa, MaMb through the points La, Lb, Lc concur at L.
4. The point L is X(376), the reflection of the centroid G with respect to the circumcenter O of ABC.

Reference:

Methods of Problem Solving. Book 2.
J. B. Tabov & P. J. Taylor
AMT Publishing