finding the center of a circle (elementary geometry) I am considering the problem of finding the center of a given circle C of radius r. I currently know 3 elementary ways (ruler+compass construction) of doing this :
- choose 2 points on the circle, draw the bisector, and draw another bisector involving a third point on the circle
- choose 2 points on the circle, draw the bisector, get a diameter then find the middle point
- choose 2 points on the circle, draw a chord, then a perpendicular line and get a diameter (Thales theorem).
I was looking for other elementary method (not involving trigonometry/analytic geometry). Then I heard something quite confusing for me : with two points on the circle C, call them A and B and "close enough", draw the circle C1 of center A and radius AB, and the circle C2 of center B and radius AB, they met at D('inner' wrt the circle C) and E('outer' wrt the circle C). Call F the other intersection of the circles C and C1, then draw (FD) which intersects the circle C at G; it seems that the circle of center G and radius r contains the three points: B, D and the center of the circle C. 
I have no idea if this is possible to prove this property using only elementary things. Any hint please? Comments are welcome (and please excuse my not-so-good english :/ )

Filling in on the main answer.  It takes an argument to show that the triangle BGD is isosceles.  We can't apply any usual triangle congruence theorem that I can see.
Here is one approach: take $A, B$, and $D$ as above.  Construct a line through $A$ which bisects $BD$.  Let the point where it crosses the circle be $G'$.  Note the construction forces angle $AG'B$ to equal angle $AG'D$.  Draw a line thru $G'$ and $D$ and call the point where this intersects the circle $F'$.  The arc $AF'$ equals the arc $BA$ because angle $AG'B$ equals angle $AG'F'$.  But this forces $F'=F$.  But then $G'$ is the point where the circle meets the line through $F$ and $D$.  So $G'=G$.
Perhaps there is a simpler way.
