- Apr 26, 2014The circumcircles of AJbAJc, BJcJ, CJJb, ABC are concurrent.J of A*B*C* instead of the a-excenter Ja. ieIf the triangle ABC is not acute, withÂ A > 90 d., then we take the incenter[APH]:B* = the other than P intersection of the circles (C', C'P) and (A', A'P)A* = the other than P intersection of the circles (B', B'P) and (C', C'P)Denote:Let ABC be an acute triangle, P a point and A'B'C' the pedal triangle of P.

C* = the other than P intersection of the circles (A', A'P) and (B', B'P)In this configuration let Ja,Jb,Jc be the excenters of A*B*C*.For P = O, The triangles ABC, JaJbJc are perspective at N.Locus of P such that ABC, JaJbJc are perspective?Also, for P = O, the circumcircles of AJbJc, BJcJa, CJaJb, ABC are concurrent.And the circumcircles of JaBC, JbCA, JcAB too.APH

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Generalization:Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P.

Denote:

A* = the other than P intersection of the circles (B', B'P) and (C', C'P)

B* = the other than P intersection of the circles (C', C'P) and (A', A'P)

C* = the other than P intersection of the circles (A', A'P) and (B', B'P)Let PaPbPc be the antipedal triangle of P wrt triangle A*B*C*The circumcircles of APbPc, BPcPa, CPaPb, ABC and also the circumcircles of

PaBC, PbCA, PcAB are concurrent.Antreas - << Previous post in topic