By famous Mohr-Mascheroni theorem
Every geometric construction that can be carried out by compass and straightedge can be done with the compass only (without a straightedge).
To say in short, to prove the theorem, we have to prove that the three following constructions can be done with only compass:
- Points of intersection of two circles given by center and one of the points for each circle
- Points of intersection of a circle (given by center and one of its points) and a straight line (given by two points).
- Point of intersection of two straight lines each of them given by two points.
I was reading "A short elementery proof of the Mohr-Mascheroni Theorem" by Norbert Hungerbuhler.
But it seems to me that the autor uses transport of the measure by the compass.
I suspect that we can avoid the usage of transport of measure by compass in the proof of Mohr-Mascheroni theorem. That is I do believe that every point constructible by collapsing compass and a straightedge can be constructed by means of collapsing compass only. But unfortunately I still find myself unable to do that.
P.S. It seems to me that despite the comments below, the construction in the Problem 4 of the book of Kostovskiy mentioned in the answer by @saltandpepper uses the measure tramsport as well [constructing the circles $(O,a)$, $(C, OE)$, $(D, OE)$].
The obstacle that bound me can be elininated by the elegant equivalence between two kinds of compasses a "rigid" one and a collapsing "divider".
Suppose we want to build a circle with center $A$ and the radius $BC$. Then we do the following: first draw blue circles, then red ones, then resulting green one; as it is shown below: [source: Wikipedia] (https://i.stack.imgur.com/aA1CA.png) Source of the answer is the underrated comment on my other old quaestion: https://math.stackexchange.com/a/2933016/239005