Dear Alxei

I wrote :

> Here is a better presentation of a generalization :

> Starting with a point J in the plane of ABC, the perpendicular

lines

> through J to JA, JB, JC intersect BC, CA, AB respectively at A',

B', C'.

> It is well known that A',B',C' are on a same line L(J).

> For any point M,

> Ma is the point of the line AM such as <MJA = <AJMa

> Mb is the point of the line BM such as <MJB = <BJMb

> Mc is the point of the line CM such as <MJC = <CJMc

> Then Ma,Mb,Mc are collinear if and only if M lies on L(J)

> In this case, the four lines A'Ma, B'Mb, C'Mc, MaMbMc touch the

> inscribed conic with focus J

> This can be proved with projective tools using the fact that M->Ma

is

> the harmonic homology with center A and axis the line JA'

> * The harmonic homology (I don't know if this English name is

correct)

> with center O, axis L maps M to his harmonic conjugate wrt O and OM

> inter L

>

>

> I think that, if we consider three harmonic homologies with centers

> A,B,C, the locus of M with three colinear images is generally a

cubic,

> even when the three axis concur. In the particular case of three

axis

> La,Lb,Lc going through a point J and such as there exists an

involution

> swapping (JA,La), (JB,Lb),(JC,Lc), then the cubic degenerates in

three

> lines :

> the fixed lines of the involution and the line through La inter BC,

Lb

> inter CA, Lc inter AB

Now, if we consider a line L with trilinear pole P and intersecting

the sidelines of ABC at A',B',C', we get :

Suppose that M is a point of L; the harmonic homology with center A,

axis JA' maps M to Ma,.., similarly Mb,Mc

Then

Ma, Mb, Mc are collinear

the four lines A'Ma, B'Mb, C'Mc, MaMbMc touch the inscribed conic

with perspector the cevian product J*P.

Looking at that, I've realized something which should be interesting

for Bernard :

For any point J, the cevian product J*o(J), where o(J) is the

orthocorrespondant of J, is the perspector of the inscribed conic

with focus J.

Friendly. Jean-Pierre