Proof that bisecting a line segment with straightedge is impossible

Question

This is the proof I read from here. I will quote it fully:

The answer is NO. To see why, consider a line L in the plane P, and two marked points A, B on it. It is desired to construct the midpoint M of the segment AB using the straightedge. Suppose we have found a procedure which works. Now, suppose we have a one-to-one mapping of plane P onto another plane P' which carries lines to lines, but which does not preserve the relation "M is the midpoint of the segment AB", in other words A, M, B are carried to points A', M', B' with A'M' unequal to B'M'. Then, this leads to a contradiction, because the construction of the midpoint in the plane P induces a construction in P' which also would have to lead to the midpoint of A'B'. (This is a profound insight, an "Aha" experience, and worth investing lots of time and energy in thinking it through carefully!!)

I don't understand how the one-to-one mapping induces an equivalent construction of the midpoint of A'B' in the plane P', given that it only preserves lines?.

Answer 1

If you’re limited to a straightedge, every step of your construction in $P$ must be drawing the line through some specified pair of points. Each of these lines is carried by the mapping to the line of $P'$ through the images of those points. Thus, the whole construction is carried to a construction that yields $M'$ from $A'$ and $B'$ instead of their actual midpoint. Yet it’s the same construction, in the sense that the points used to determine each successive line of the construction are determined by the same rule applied to $A'$ and $B'$, so if it really worked, it would have to construct the actual midpoint of $\overline{A'B'}$.

Answer 2

It depends on your "initial position". Given A, B, and possibly other points on the line AB (but not M) we cannot obtain other points on AB unless we are given at least 2 points C,D not on AB. Now suppose M lies on the line CD. If we have a projection ,that is, a perspective view P' of P, the recipe "Take the intersection M of CD with AB" transforms to "Take the intersection of A'B' with C'D' " which may or may not give the mid-point of A'B' because projections in general do not preserve ratios of lengths and do not in general preserve mid-points,, but M is still the mid-point of AB . I don't know which book you refer to.I don't intend to read it.