Re: [EMHL] Feuerbach
- Dear Darij, I wrote
> For any point P, let F = common point of the NPC of ABC, PBC, PCA,Of course, this result is slightly different if ABCP is cyclic:
> PAB = center of the rectangular circumhyperbola going through P
> Let N, Na, Nb, Nc be the NP-centers of ABC, BCP, CAP, ABP.
> Q = isogonal conjugate of P, QaQbQc= pedal triangle of Q.
> Then the figures QQaQbQc and NNaNbNc are indirectly similar.
> Moreover, if Q' = inverse of Q in the circumcircle = isogonal
> conjugate of the reflection of P wrt F, NaNbNc and the pedal
> triangle of Q' are homothetic; the ratio is OQ^2/(OQ^2-R^2).
F is the midpoint of PH; NNaNbNc lie on the circle (F,R/2) and the
triangle NaNbNc is translated from the medial triangle by
NF = 1/2 OP.