- Mar 25 12:07 AM[APH]:[Angel ]:The lines BcCb, CaAc, AbBa are concurrent.Ac = (perpendicular from A' to BC) /\ PCAb = (perpendicular from A' to BC) /\ PBDenote:Let ABC be a triangle, P = I and A'B'C' the antipedaltriangle of P.

Similarly:

Bc = (perpendicular from B' to CA) /\ PC

Ba = (perpendicular from B' to CA) /\ PA

Ca = (perpendicular from C' to AB) /\ PA

Cb = (perpendicular from C' to AB) /\ PBPoint?In general, the locus of P such that the lines BcCb, CaAc, AbBa areconcurrent is Darboux cubic.For every point P the lines BcCb, CaAc and AbBa are concurrent.If P = (x: y: z), then the point of concurencia of BcCb, CaAc and AbBa is Q=(x (SB y^2 + SC z^2 + SA (y + z)^2):...:...)Some pairs {P, Q}:{1, 8}, {2, 599}, {3, 3}, {4, 3}, {6, 599}, {8, 2098}, {13, 298},{14, 299}, {15, 298}, {16, 299}, {36, 4511}, {56, 2098}, {80, 4511}, ...

***************************************************Dear AngelWhenever the locus is the entire plane, the concurrence can be

proved synthetically.In our case we have:1. BaCa, CbAb, BcAc are concurrent at P.2. AbAc, BcBa, CaCb are concurrent, since ABC, A'B'C'are orthologic by conctruction, so they concur at the otherthan P orthologic center.Now by the Double Perspectivity Theorem we get that

BcCb, CaAc, AbBa are concurrent.APH - << Previous post in topic Next post in topic >>