Dear Darij,

I've tried to find an easy synthetic proof of Braun theorem

> The Miquel point configuration is defined as follows:

>

> On the sidelines BC, CA, AB of a triangle ABC, points A', B', C'

are

> chosen and the circles AB'C', BC'A', CA'B' are drawn. Then, as we

> know, these circles have a common point P (the Miquel point).

>

> Now the two theorems:

>

> 2. Let Ma, Mb, Mc be the centers of the circles AB'C', BC'A',

CA'B'

> [it is well-known that triangles MaMbMc and ABC are similar], let

M'

> be the circumcenter of triangle MaMbMc, and M the circumcenter of

> triangle ABC. Then M'M = M'P.

>

> [This is a theorem by Peter Baum; proposed in the little German

> periodical "Die Wurzel" (see the website wurzel.org), in the 12/98

> issue. Solved by Sefket Arslanagic in the 5/99 issue; the solution

> was quite involved.]

PaPbPc is the pedal triangle of P.

A direct similitude with center P, angle T, ratio 1/cos(T) maps

PaPbPc to A'B'C'.

As the circumcenter of PPbPc is the midpoint A'' of AP, the

similitude maps this midpoint to Ma; hence the similitude maps the

midpoint I of PM (circumcenter of A''B''C'') to M'(circumcenter of

MaMbMc).

As PM'/PI = cos(T) and <IPM' = T, it follows that M' lies on the

perpendicular bisector of PM.

Friendly. Jean-Pierre