- Dear Ricardo, Could you upload any proof of the Hagge's theorems you show us?

Thanks

José Carlos

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[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

--- 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 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

Nikos Dergiades

> Dear Ricardo, Could you upload any proof of the Hagge's theorems

> you show us?

>

> Thanks

>

> José Carlos

> - A correction please:

I wrote

> We know that with signed segments we have

the correct is

> HA.HH1 = HB.HH2 = HC.HH3 = -2cosA.cosB.cosC

HA.HH1 = HB.HH2 = HC.HH3 = -4R^2cosA.cosB.cosC

Best regards

Nikos Dergiades