(4)... circumcenter of triangles BPC,CPA,APB. >Let A',B',C' be midpoints of AOa,BOb,COc then NPC center of A'B'C' lies on Euler line of ABC. Denote Np = the NPC
(4)Let ABC be a triangle and I1,I2,I3 the reflections of I in BC,CA,AB, resp. Denote: I12, I13 = the orthogonal projections of I1 on AC,AB, resp. I23, I21 = the
(3)Let ABC be a triangle, P a point and PaPbPc the antipedal triangle of P. Denote: AaAbAc, BaBbBc, CaCbCc = the pedal triangles of Pa, Pb, Pc, resp. A'B'C.' =
Fetching Sponsored Content...
(14)Orthologic centers and homography Angel Montesdeoca, Centros ortológicos y homografía ...
(5)[A variation of a configuration by APH] Let ABC be a triangle, A'B'C' a variable triangle w/r to ABC and A1B1C1, A2B2C2 the medial triangles of ABC, A'B'C',
(5)[APH]: Let ABC be a triangle and P a point. Denote: Pa = (Reflection of BC in PB) /\ (Reflection of BC in PC) Pb = (Reflection of CA in PC) /\ (Reflection of
(6)On Wed, Mar 4, 2015 at 7:11 PM, Antreas Hatzipolakis wrote: Let ABC be a triangle. ... ************************ On Wed, Mar 4, 2015 at 8:51 PM, Moses, Peter J.
(3)1. La, Lb, Lc are concurrent? (the entire plane, with point of concurrence the isogonal conjugate of P) Confirmed 2. ABC, A'B'C' are perspective? (the entire
(4)[APH]: The perpendiculars to HIa, HIb, HIc from A,B,C, resp. bound triangle A"B"C". Which point is the circumenter of A"B"C" ? Is it lying on the IH line? ...
View First Topic Go to
Loading 1 - 9 of total 9 topics