How do we find the mid-point of a line segment using only a compass?

I believe there is a theorem that says you can do all Euclidean
constructions with only a compass, and also that you can do all
Euclidean constructions with only a straightedge provided that you have
the center and an arc of a circle anywhere in the plane.  I do not know
where to find these theorems.

Problem: Given line segment ab,

                                a=======b

construct the point that bisects the line using only a compass (no
straightedge).

Solution: Construct the sequence of points c,d,e,f,g,h,i by completing
the equilateral triangles abc, acd, ade, def, efg, fgh, and ghi:

            h-------f-------d-------c
           / \     / \     / \     / \
          /   \   /   \   /   \   /   \
         /     \ /     \ /     \ /     \
        i-------g-------e-------a=======b

(You can find the points, but, of course, you can't draw the lines
segments connecting them since you don't have a straightedge.)

Now, find an intersection, j, of the circle with center i and radius ie
and the circle with center e with radius ea (forgive me for not drawing
the circles in ascii).

                      j
            h-------f-------d-------c
           / \     / \     / \     / \
          /   \   /   \   /   \   /   \
         /     \ /     \ /     \ /     \
        i-------g-------e-------a=======b

With point j as the center and radius ji, draw the arc that intersects
ab.

                      j
            h-------f-------d-------c
           / \     / \     / \     / \
          /   \   /   \   /   \   /   \
         /     \ /     \ /     \ /     \
        i-------g-------e-------a===k===b

Assertion: The point of intersection, k, bisects ab.

Proof:  The proof is relatively easy to do with analytic geometry, so
without loss of generality, we place a coordinate system around the
line segment ab such that a is at (0,0) and b is at (1,0).  We proceed
to prove that the point k is at (1/2,0).

Since the triangles are equilateral, it follows that igeab is a
straight line.  Point e is at (-1,0), point g is at (-2,0), and point i
is at (-3,0).  You can verify that the point at
(-5/4,sqrt(4-(7/4)^2)) ~= (-1.25,.968...) is 2 units from point i and 1
unit from point e.  This point therefore is the point j.  (Obviously
there is another intersection point j' at (-5/4,-sqrt(4-(7/4)^2)); you
can choose either one.)

Finally, the circle with center j and radius ji = 2 intersects line ab
at point (x,0) such that

        (-5/4-x)^2 + sqrt(4-(7/4)^2)^2 = 2^2 .

Simplifying, we get

        (-5/4-x)^2 + 4 - (7/4)^2 = 4
and
        (-5/4-x)^2 = (7/4)^2

Taking square roots,

        -5/4-x = +or- 7/4
or
        x = +or- 7/4 - 5/4

yielding the solutions x=1/2 and x=-3.  Thus, point k is at (1/2,0),
and therefore is the bisector of line segment ab.  QED

Even if the line ab is not drawn, it still is possible to find the
point that bisects ab.  Find both points j and j' as described above.
Then find point k as the intersection of arcs centered at j and j' with
radii ji = j'i.  By symmetry, k is on ab, and the above proof shows
that it bisects ab.

                      j
            h-------f-------d-------c
           / \     / \     / \     / \
          /   \   /   \   /   \   /   \
         /     \ /     \ /     \ /     \
        i-------g-------e-------a---k---b




                      j'

All of the above constructions can be done even with a Euclidean
compass, which collapses when you lift it up so that you cannot use it
to mark the distance between two points, lift it, and use it to copy
the distance.

-- Dave Dodson <dodson@convex.COM>