Dear friends

I wrote

> Here is a more symetric way to draw this figure :

> Ca is the circle going through A with center the common point of

the

> perpendicular bisector of BC with the line AX[74] (The radius of

this

> circle is OH/|2.cos A|). Similarly, we get Cb, Cc (may be, there is

a

> special name for these three circles because they are closely

> connected with the Mac Cay circles)

> Three parallel lines going respectively through A, B, C intersect

> again Ca, Cb, Cc respectively at A'', B'', C''; A' = BC inter

> B''C'', ...

> I've checked by computation, that, if we consider three directly

> congruent non isosceles triangles inscribed in the circumcircle,

> Ta,Tb,Tc where A is a vertex of Ta and La the opposite sideline,

then

> the points BC inter La, CA inter Lb, AB inter Lc will lie on a same

> line only in the case above.

In fact, this is a bit tedious because the centers of the three

circles Ca, Cb, Cc are homothetic in (O,-1/2) of the vertices of the

tangential triangle.

That's enough for me for today.

Friendly. Jean-Pierre