Answers are grounded in this paper and cite their sources.
Let A1B1C1 be the Cevian triangle of a point K = (p:q:r), cutting off three corner triangles from ABC. For a point P = (u:v:w), barycentric weights applied to the corner triangles under opposite and direct vertex orientations give two companion triangles, the anti-homothetic corner associate triangle PA PB PC and the direct corner associate triangle QA QB QC. We prove that both are always perspective with ABC, with explicit perspectors, and that their degeneracy loci are a circumconic and a pair of lines respectively. The lines PA QA, PB QB, PC QC are parallel to the sides of the Cevian triangle and concur exactly when P lies on a canonical line. For P = K, the triangle KA KB KC has the same area as the Cevian triangle. For K = H (orthocenter), PA, PB, PC are the similarity images of P in the three orthic corner triangles and [PA PB PC]/[ABC] = (R^2 - OP^2)/(4R^2); we derive sharp (R, r, s) inequalities. Preprint of a manuscript submitted for publication (submitted version, 5 November 2026).
No takes yet. Share an insight, caveat, or question.
Hoang Long Nguyen (2026) studied this question.
Synapse has enriched 8 closely related papers on similar clinical questions. Consider them for comparative context: