## [EMHL] Hagge Theorem (0, 1, 2, 3)

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• Dear Ricardo, Could you upload any proof of the Hagge s theorems you show us? Thanks José Carlos ... LLama Gratis a cualquier PC del Mundo. Llamadas a fijos y
Message 1 of 4 , Feb 28, 2006
Dear Ricardo, Could you upload any proof of the Hagge's theorems you show us?

Thanks

José Carlos

---------------------------------

LLama Gratis a cualquier PC del Mundo.
Llamadas a fijos y móviles desde 1 céntimo por minuto.
http://es.voice.yahoo.com

[Non-text portions of this message have been removed]
• Dear José Carlos, and all: Jonhson proposes them like exercises. Only them I have seen with Cabri. Ricardo ...
Message 2 of 4 , Feb 28, 2006
Dear José Carlos, and all:

Jonhson proposes them like exercises.

Only them "I have seen" with Cabri.

Ricardo

--- José Carlos Chávez Sandoval <hucht_sc@...>
escribió:

> Dear Ricardo, Could you upload any proof of the Hagge's
> theorems you show us?
>
> Thanks
>
> José Carlos
>
>
> ---------------------------------
>
> LLama Gratis a cualquier PC del Mundo.
> Llamadas a fijos y móviles desde 1 céntimo por minuto.
> http://es.voice.yahoo.com
>
> [Non-text portions of this message have been removed]
>
>

______________________________________________
LLama Gratis a cualquier PC del Mundo.
Llamadas a fijos y móviles desde 1 céntimo por minuto.
http://es.voice.yahoo.com
• Dear José Carlos, Let ABC be a triangle H its orthocenter H1H2H3 its cevian triangle. Let P be an arbitrary point with cevian triangle A B C . We know that
Message 3 of 4 , Feb 28, 2006
Dear José Carlos,

Let ABC be a triangle H its orthocenter H1H2H3 its cevian triangle.
Let P be an arbitrary point with cevian triangle A'B'C'.
We know that with signed segments we have
HA.HH1 = HB.HH2 = HC.HH3 = -2cosA.cosB.cosC

1) Let the perpendicular from H meets the circle with diameter AA'
at the points A1,A2 and similarly define on the circles (BB'), (CC')
the points B1, B2, C1, C2.
Since the points A, H1, A1, A2 are concyclic then
HA1.HA2 = HA.HH1 similarly
HB1.HB2 = HB.HH2
HC1.HC2 = HC.HH3.
Hence HA1.HA2 = HB1.HB2
the points A1, A2, B1, B2 are concyclic and the
center of the circle lies on the lines AA', BB' and
this is the point P.
Similarly the points A1, A2, C1, C2 are lying on a
circle with center P that is same as above.

2) The circle (BC) passes through H2, H3
meets the line HA1A2 at the points A3, A4.
Similarly define the points B3, B4, C3, C4.
Hence HA3.HA4 = HB.HH2
Thinking as in (1) we conclude that
the points A3, A4, B3, B4 are concyclic and the
center of the circle lies on the line from the mid point
of BC parallel to AA' and on the line parallel from the
mid point of CA to BB'. This point Q is the complement of P.
(perhaps there is an error in Jonson I think.
The phrase
whose center, with reference to triangle A1 A2 A3 is
homologous to the position of P in the triangle A1A2A3
must be
whose center, with reference to medial triangle of
A1A2A3 is homologous to the position of P in the triangle A1A2A3)
(A1A2A3 in Jonson is the triangle ABC)

3) The circle (BC) passes through H2, H3
meets the circle (AA') at A5, A6 and
since HA.HH1 = HB.HH2 the point H lies
on the radical axis of the circles (BC), (AA').
Hence H lies on the line A5A6 and we have
HA5.HA6 = HA.HH1 = HB5.HB6 = HC5.HC6
Similarly thinking as in (1) and (2)
the points are concyclic and the center of the
circle is the point that lies on the lines
AoA", BoB", CoC" where AoBoCo is the medial
triangle of ABC and A" is the intersection of BoCo
with the line AP. Similarly define B", C".
This center is the isotomic conjugate of Q wrt
the medial triangle AoBoCo.

Best regards

> Dear Ricardo, Could you upload any proof of the Hagge's theorems
> you show us?
>
> Thanks
>
> José Carlos
>
• A correction please: I wrote ... the correct is HA.HH1 = HB.HH2 = HC.HH3 = -4R^2cosA.cosB.cosC Best regards Nikos Dergiades
Message 4 of 4 , Feb 28, 2006