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• Dear Antreas ... The common points of the NP-circles of PaAB and PaAc are A and the midpoint of APa. Hence, the line NabNac is the perpendicular bisector of
Message 1 of 2 , Jan 1, 2004
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Dear Antreas
> Let ABC be a triangle, A'B'C' its orthic triangle,
> P a point and Pa, Pb, Pc the orthogonal projections
> of P on AA',BB',CC', resp.
>
> Denote
>
> Nab = The Nine Point Circle Center of PaAB
> Nac = The Nine Point Circle Center of PaAC
>
> Nbc = The Nine Point Circle Center of PbBC
> Nba = The Nine Point Circle Center of PbBA
>
> Nca = The Nine Point Circle Center of PcCA
> Ncb = The Nine Point Circle Center of PcCB
>
> The "diacentral" lines NabNac, NbcNba, NcaNcb
> are concurrent parallelians (ie parallels to the
> sidelines of ABC)
>
> What is the point of concurrence?
> [Special case: P on the Euler Line]

The common points of the NP-circles of PaAB and PaAc are A' and the
midpoint of APa. Hence, the line NabNac is the perpendicular
bisector of the segment joining these two common points.
It follows easily that your common point is the homothetic of P in
(G,1/4).
Friendly. Jean-Pierre
• Let ABC be a triangle, A B C its orthic triangle, P a point and Pa, Pb, Pc the orthogonal projections of P on AA ,BB ,CC , resp. Denote Nab = The Nine Point
Message 2 of 2 , Jan 1, 2004
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Let ABC be a triangle, A'B'C' its orthic triangle,
P a point and Pa, Pb, Pc the orthogonal projections
of P on AA',BB',CC', resp.

Denote

Nab = The Nine Point Circle Center of PaAB
Nac = The Nine Point Circle Center of PaAC

Nbc = The Nine Point Circle Center of PbBC
Nba = The Nine Point Circle Center of PbBA

Nca = The Nine Point Circle Center of PcCA
Ncb = The Nine Point Circle Center of PcCB

The "diacentral" lines NabNac, NbcNba, NcaNcb
are concurrent parallelians (ie parallels to the
sidelines of ABC)

What is the point of concurrence?
[Special case: P on the Euler Line]

Happy New Year

Greetings from Athens

Antreas

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