Enclosing rectangles

If we construct enclosing rectangles BCCaBa, CAAbCb, ABBcAc to the triangle ABC, then the lines AbBa, CaAc and BcCb make a triangle perspective with ABC. The perspector is catalogued as X(66) in Kimberling's ETC. This point is the isogonal conjugate of Exeter point, X(22).

Generalization (Nikolaos Dergiades, Hyacinthos message #15778): Let XYZ and DEF be the orthic and medial triangle of ABC, respectively.

  • The line BcCb is the reflection of line YZ with respect to D,
  • The line AcCa is the reflection of line ZX with respect to E,
  • The line AbBa is the reflection of line XY with respect to F,

hence the we can generalize the problem considering any point P=(u:v:w) and its cevian triangle XYZ. For, we have that

  • The reflection of line YZ with respect to D is the line ra: (u(v+w)+vw) x + uv y + uw z = 0,
  • The reflection of line ZX with respect to E is the line rb: uv x +(v(w+u) + wu)y + vw z =0,
  • The reflection of line XY with respect to F is the line rc: uw x + v w y + (w(u+v)+uv) z =0,

The lines rb an rc intersect at the point

A' = (-(v + w) (u v + uw + 2 v w): (v(uv + uw + vw - w^2)): (w(uv - v^2 + uw + vw))),

and similarly we get the points B', C'. The lines AA', BB', CC' intersect at

Q= (u/(u(v+w)+vw-u^2) : v/(v(w+u)+wu-v^2) : w/(w(u+v)+uv-w^2)).

 
Special case: Since the sum of homogeneus barycentric coordinates of Q can be factored as uvw(uv+vw+wu), we get that Q is an infinity point if P is on the Steiner circumellipse.