Constructing the midpoint of a segment by compass When I am working with my child, I am stuck in this geometry problem.
"We have two different points $M, N$ in the plane. Using only compass to construct the midpoint $I$ of the segment $MN$."
Thank you for all helping and comments.
 A: Some Googling revealed the following comments to this answer:

  
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*I know it is possible, but is there an easy way to divide a segment in half with only a compass? – robjohn♦ May 20 at 3:46
  
*I don't know if that's "easy", but here's one method:
  
  
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*Find the point $C$ on the ray from $A$ through $B$ such that $|AC|=2|AB|$ using my previous comment [The relevant part: "To double the distance along a ray, use the construction of a regular hexagon with vertex $A$ and center $B$".]
  
*Intersect the circle with center $C$ through $A$ with the circle with center $A$ through $B$ to find $D_1,D_2$.
  
*The midpoint of $AB$ is the second point of intersection of the two circles with center $D_i$ through $A$. – t.b. May 20 at 9:28
  
  
*Here is a picture of what I have in mind: - t.b. May 20 at 12:38
  


The dotted line is not used in the construction.
Added:
The triangles $\Delta ACD_1$ and $\Delta AMD_1$ are isosceles by construction and they share a common angle, hence they are similar. Therefore $AM : AB = AM : AD_1 = AD_1 : AC = AB : AC = 1 : 2$.
A: Open the compass to any length more than half the distance between $\,A,B\,$ but less than their total distance. Put the compass's point on A and trace part of the circle over the line $\,AB\,$ , and after this do the same putting the point on $\,B\,$, (without changing the compass's openning!) and mark the interesection point of the two circles as $\,P\,$.
Now repeat the above with circles under the line segment and mark the intersection point of the two circles as $\,S\,$ (BTW, no need the compass has the very same openning as in the first part!). Since both points $\,P,S\,$ are at the same distance from $\,A\,$ and from $\,B\,$ (why?) , joining them gives you the perpendicular bisector of $\,AB\,$.
Finally, just take the intersection of the Perp. bisector with the segment $\,AB\,$
A: Finally, by google we find the solution on thess sites:
http://gogeometry.com/circle/mascheroni_compass_1.htm
http://mathafou.free.fr/themes_en/compas.html
and useful lecture on this problem
http://www.math.ualberta.ca/~tlewis/343_10/04sec.pdf
