- GENERALIZATION

Let ABC be a triangle, Q a point, QaQbQc the pedal triangle of Q,

(X) the circumcircle of QaQbQc (=pedal circle of Q), P a point on

the OQ line and PaPbPc the pedal triangle of P.

Denote:

(X1), (X2), (X3) the reflections of (X) in OPa,OPb,OPc, resp.

A'B'C' = the triangle bounded by the radical axes of ((O),(X1)),((O),(X2)),((O),(X3))

Conjecture:

The triangles A'B'C', X1X2X3 are perspective.

Figure:

http://anthrakitis.blogspot.gr/2013/02/reflecting-pedal-circle.html

Antreas

[APH]>

> Denote:

>

> PaPbPc = the pedal triangle of a point P on the Euler line

>

> (N1), (N2), (N3) = the reflections of (N) in PaO, PbO, PcO, resp.

>

> A'B'C' = the triangle bounded by the radical axes of

> ((O),(N1)), ((O),(N2)),((O),(N3)).

>

> A'B'C', N1N2N3 are perspective.

>

> True??? - Are the triangles ABC, A'B'C' perspective?Feuerbach point Fc.The NPC of CHIc intersects the excircle (Ic) at C' other than theFeuerbach point Fb,The NPC of BHIb intersects the excircle (Ib) at B' other than theFeuerbach point Fa.Let ABC be a triangle with excenters Ia,Ib,Ic.

The NPC of AHIa interscts the excircle (Ia) at A' other than theIn any case, has the triangle A'B'C' any interesting properties?APH