Antreas wrote

> Your theorem rephrased:

>

> The perpendiculars from each vertex of the antipedal triangle of I

> to the line joining the Feuerbach point with the corresponding

vertex

> of the pedal triangle of O are concurrent.

>

> The "dual" (by replacing the points I,O):

>

> The perpendiculars from each vertex of the antipedal triangle of O

> to the line joining the Feuerbach point with the corresponding

vertex

> of the pedal triangle of I are concurrent.

>

> Is it true??

>

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

I don't think so (if I may believe my sketches)

I also tried anticevian triangles instead of antipedal triangles

(because IaIbIc is the antipedal and anticevian triangle of I) but

it doesn't work either

I also tried replacing one of the excenters by the incenter and

replacing the Feuerbach point by the tangent point of the

ninepointcircle and the selected excircle but without success

PS

I would like to thank Paul for the coordinates of this point

and Jean-Pierre for some nice properties of it and I'm sorry for

the typing errors in the formulae in my original mail

Greetings from Bruges

Eric Danneels