Σάββατο 15 Αυγούστου 2026

A CYCLOLOGIC THEOREM RELATED TO ORTHIC TRIANGLE

[APH]

Orthic Version

Let ABC be a triangle, HaHbHc the orthic triangle and P a point.

Denote:

Pa, Pb, Pc = same to P points of AHbHc, BHcHa, CHaHb, resp.

HaHbHc, PaPbPc are circumcyclologic

Cyclologic center (HaHbHc, PaPbPc = ? (on the circumcircle of HaHbHc = NPC)
Cyclologic center (PaPbPc, HaHbHc) = ? (on the circumcircle of PaPbPc)

[Ercole Suppa]

cyclologic center (HaHbHc, PaPbPc) = Poncelet point(isogonal conjugate(P))

cyclologic center (PaPbPc, HaHbHc) = orthoassociate(isogonal conjugate(circumcircleInverse(P)))

Euclid 10026

-

Δεν υπάρχουν σχόλια:

Δημοσίευση σχολίου

Εγγραφή σε: Σχόλια ανάρτησης (Atom)

ETC1

X(73160) = X(2)X(6120)∩X(15859)X(66554) Barycentrics    Sin[B/3]*Sin[(2*B)/3]*Sin[B] + (Sin[A]*Sin[B] + Sin[C/3]*Sin[(2*C)/3])*Sin[C] : : ...

  • EUCLID 4404
  • ANOPOLIS CIRCLE
    Let ABC be a triangle. Denote: (N1),(N2), (N3) = the NPCs of IBC, ICA, IAB, resp. (12), (13) = the reflections of (N1) in BI, CI, resp. ...
  • HOMOTHETIC TRIANGLES and EULER LINES
    Theorem 1. Let ABC be an equilateral triangle and P a point. The Euler lines of the triangles PBC,PCA,PAB are concurent.Denote the point ...