## 22189Re: CONCURRENT CIRCLES (Re: Three circles centered at the pedals of a point - Locus

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• Apr 26, 2014
[APH]:

Let ABC be an acute 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)

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.

If the triangle ABC is not acute, withÂ  A > 90 d., then we take the incenter
J of A*B*C* instead of the a-excenter Ja. ie
The circumcircles of AJbAJc, BJcJ, CJJb, ABC are concurrent.
APH

*************************************************************************

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

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