How can we determine the center of a circle using a straightedge? Given a circle with diameter AB, how can we determine the center of the circle with a straightedge (we cannot measure lengths, cannot measure angles, or draw parallel lines,... We can only draw straight lines)?
I received this problem as homework from my professor.
In fact, we were finding a way to apply a trapezoid theorem(!) to the problem in which we have to construct a line through any point C on a circle perpendicular to its diameter AB with only a straightedge. That problem was indeed doable. The professor asked me, as an advance exercise, to try using the theorem to find the center of a circle with A and B being two given points on the circle and segment AB being the diameter.
(!) Given a trapezoid, the straight line joining the point of intersection of its diagonals and the point
of intersection of its non-parallel sides bisects each of the parallel sides.

 A: Edit: The following answer was written assuming that the ambiguous phrase “given a circle with diameter $AB$” in the question is to be interpreted as “given a circle having diameter $AB$” rather than “given a circle and given one of its diameters $AB$”.  Assuming the latter interpretation, the argument can be modified by considering the isometries of the hyperbolic plane (represented through its Klein model) which fix two ideal points, where the circle is seen as the set of ideal points.
It is a classical theorem due to Jacob Steiner and David Hilbert that it is not possible to construct the center of a circle with straightedge alone: the classical argument is essentially that stated by Fedor Petrov in his answer; see Cut the Knot for an explanation, and Shen - Hilbert's error? for a criticism of the way this argument is (sometimes carelessly) formulated.
Of related interest is the Poncelet–Steiner theorem which states essentially that given a single circle and its center anywhere in the plane, any compass-and-straightedge construction can be performed with straightedge only.
A: I do not think that it is possible.
Consider a projective map $f$ which preserves the circle; maps the center $O$ to a point $P\ne O$; and maps some point $Q$ to $O$. The line $QO$ is mapped to the line $OP$. So, there is a rotation $g$ with respect to $O$ such that $g(f)$ maps the line $QO$ to itself, although $O$ not to itself. It yields that given a circle and a line through $O$, you can not find the center by using straightedge only, since such construction is projective-equivariant.
A: Please refer to Chapter 18 of "The Ruler in Geometrical Constructions" by Smogorzhevskii. There the author provides a provides a proof of sorts that in general where you have two(!) circles, you can't construct their midpoints using only a ruler. If you can't do it in general with two, you certainly can't do it with a single given circle and only a ruler.
There are cases where you can construct the midpoints of two circles, and one of these is two given intersecting circles.  See chapter 19 of the same book.
A: Coming at this from a total non-mathematician (found this on HNQ). So no fancy proofs (last did any of those in the mid-1980s).

*

*Do I start with the diagram (circle with one line already marked)? (Not clear, but some comments seem to indicate that is the case.)


*Can I mark things ("only draw straight lines" - but can I draw some short straight lines on the straightedge?)?
If both are true then:

*

*Put the straightedge across AB

*Mark A and B on the straightedge

*Move the straightedge around (angle doesn't matter) and put the A mark on the edge of the circle (we'll call this spot C)

*Rotate the straightedge until the B mark is on the opposite side of the circle (we'll call this spot D)

*Draw a line CD

*The intersection of AB and CD is the center

Or am I vastly oversimplifying or cheating or somehow missing the point? (And if any of those are going on, I will delete this answer.)
