Dear Eisso!

>

>* Your properties 1 & 2 give rise to an easy and/or elegant

>construction of a bi-centric quadrilateral, something I have been

>looking for for a while. All you need to do is to take a circle and a

>point P inside the circle. The four points of intersections of any pair

>of perpendicular lines through with the circle will form a cyclic

>quadrilateral and the KLMN that you construct will be bi-centric. When

>you construct a bi-centric KLMN in this way, it turns out that both

>``centers'' of KLMN and the center on the circumscribed circle of ABCD

>are collinear. In fact, the circumcenter to KLMN lies exactly halfway

>between the incenter of KLMN and the circumcenter of ABCD. Perhaps this

>property can be used to give an alternative proof of Krafft's (?)

>formula for the relation between the radii of incenter and excenter of

>a bicentric quadrilateral and the distance between the centers of the

>circles.

>

You are right. This proof of Ponsele theorem for n=4 is typed in Sharygin's

book "Problems of planymetry".

Sincerely Alexey