Thanks Antreas.

[ND]

> >it is known http://mathworld.wolfram.com/AdamsCircle.html

> >that the parallel lines from the Gergonne point

> >Ge = ( 1/(s-a) , 1/(s-b), 1/(s-c) ) in barycentrics

> >of triangle ABC to the sides of its cevian triangle meet the

> >sides of ABC at six points that are lying on Adams' circle

> >a circle with center the incenter of ABC.

> >

> >Is it known that the same property holds also for the other

> >Gergonne points Ga, Gb, Gc of triangle ABC?

[APH]

> This should be known by Conway's extraversion

>

> BTW, and Conway's Circle is also centered at I and has

> three brother-circles by extraversion.

If the point P has barycentrics (x : y : z)

and A'B'C' is its cevian triangle then the parallel line

from P to B'C' meets the sides AC, AB of triangle ABC

at the points Ab, Ac. Similarly define the points Bc,Ba, Ca, Cb.

The six points are lying on a conic with center the point

( x(y+z) : y(z+x) : z(x+y) ) in barycentrics that is the

complement of the isotomic conjugate of P.

If the point P is inside the Steiner circumellipse the conic

is an ellipse or circle for the points Go, Ga, Gb, Gc.

If the point P is on the Steiner circumellipse the conic

is a parabola.

If the point P is outside of the Steiner circumellipse the conic

is an hyperbola and if P is also on the conic

xx + yy + zz + 3xy + 3yz + 3zx = 0

the Steiner circumellipse of the anticomplement triangle of ABC

the points Ab, Bc, Ca are collinear and the points Ac, Ba, Cb

are also collinear.

Best regards

Nikolaos Dergiades