I'm aware of counter examples to Playfair's axiom-"Given a line L and a point P not on that line, there is one and only one line passing through point P which is parallel to L." These are violated in spherical geometry where all lines intersect so there are no parallel lines and in hyperbolic geometry where there are multiple parallel lines through a point.

I'm unaware of any exceptions WITHIN the plane. Euclid avoids using the parallel postulate for almost the first 30 propositions in The Elements. Others have made attempts to prove the parallel postulate only for it not to be accepted despite the counterexamples of spherical and hyperbolic geometry not having yet been discovered. What was the basis of that objection without the new geometries? Apparently proofs were rejected despite constructive examples we have today. How were they able to do that?

More specifically, without using what we know from spherical geometry and hyperbolic geometry, where does the following "theorem" fail?


Given a line L and a point C not on the line, construct a second line M through C and perpendicular to L. Also Construct a line through C perpendicular to M called N.

1) Start with line L and point C not on L. 2) Select an arbitrary point on L, Q1. Construct a circle through C centered at Q1. 3) Select another, different, arbitrary point on L,Q2. Construct a circle through C centered at Q2. 4) Draw a line, M, through C and the other intersection of these two circles, R. 5) Pick a point, P, on the same side of L as C and on the opposite of M as C. 6) Construct a circle centered at P passing through C. Call the other place this circle intersects M, M1. 7) Construct a line through M1 and P, calling the farthest point of intersection on CircleP from M1, P1. 8) Draw a line,N, through P1 and C. 9) Claim, N is parallel to L and perpendicular to M.


Proof N is perpendicular to M and M is perpendicular to L. The segment connecting Q1 and Q2 and the upper radii containing C of either circle creates a triangle congruent to the triangle fromed by the same segment, Q1Q2, and the lower radii, containing R, of either circle. These triangles being congruent, it follows that L is an angle bisector of angle CQ1R. By SAS, it follows that L is the bisector of segment RC. It also follows that the angle made by L and M is a right angle because the angles are congruent and sum to a straight line. So M is perpendicular to L. The line M1 through P to P1 is a diameter of the circle centered at P. By Thales' Theorem, the line from P1 to C , called N, is perpendicular to M.

Proof?? two lines perpendicular to a third are parallel to each other Given lines L,M, and N with M intersecting L and N at right angles, intersecting N at point C with point C not on L: Construct a circle centered at point C to an arbitrary point,X1, on L. Draw a line through C and X1. Determine the midpoint of line CX1. Draw a line from the intersection of L and M through that midpoint. Construct a circle centered at the midpoint and passing through the intersection of M and L. This circle creates a new intersection R, as far from the midpoint as the intersection of M and L. Vertical angles are congruent and we've constructing intersecting diagonals intersecting at their midpoints. The resulting triangles are therefor congruent by SAS. These triangles being congruent, it follows that the segments opposite the vertical angles are congruent. THe segment lying on M is the distance along M between L and N, so the distance in the other triangle between the points on L(point X1) and N is the same. By similar arguments, the other pair of vertical angles imply that R is as far from C as X1 is from the intersection of L and M. By SSS, it can be proven that the line between R an X1 is perpendicular to both L and N. X1 was chosen arbitrarily, so it follows that the length of a transversal perpendicular to L at X1 will intersect N a distance from X1 equal to the distance between C and the intersection of M and L. This distance is never zero therefor N and L do not intersect.

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    $\begingroup$ As I commented elsewhere, spherical geometry is totally irrelevant to this story, since it fails to satisfy more axioms of Euclidean geometry than just the parallel postulate. $\endgroup$ – Eric Wofsey Sep 20 '18 at 22:33

I have not checked all the details of your specific argument, but your general idea to construct parallel lines by constructing perpendiculars twice is valid in "neutral geometry" (Euclidean geometry minus the parallel postulate), and indeed constructs a line $N$ through $C$ which is parallel to $L$. This does not prove Playfair's axiom, though, because Playfair's axiom further claims that this line $N$ is unique. That is, to prove Playfair's axiom you would need to prove that there are no other lines through $C$ parallel to $L$ besides the one you have constructed.

As for your more general question of how people historically rejected proofs of the parallel postulate, the answer is very simple. The entire point of axiomatic geometry is that we start with a fixed list of axioms, and all of our proofs must be based only on those axioms. All of the purported proofs of the parallel postulate either simply had an error, or relied on some additional assumption that was not one of the axioms.

Note that this is still how it works today, even though we know about hyperbolic geometry. Since we know hyperbolic geometry satisfies all the axioms of neutral geometry but does not satisfy the parallel postulate, we know that any proof of the parallel postulate from neutral geometry must have a flaw somewhere. But this doesn't actually tell us what is wrong with any particular given proof; to identify that, we have to actually find the flaw in the logic (as we know there must be one somewhere).

(To be fair, the history is a little more complicated, since there were various unstated axioms that people implicitly considered valid which were not identified until the 19th century. None of these assumptions were relevant to the issues with proposed proofs of the parallel postulate, though, and these axioms that were unstated are true in hyperbolic geometry too.)


Your theorem does not fail and was a known result for much of the time that people debated the parallel postulate: in fact, it is Proposition I.31 in Euclid's axioms. Euclid's proof is much the same as yours, except that it is not required to take the lines to be perpendicular. It's enough to copy an angle, getting one of those pictures you see many times in a high school geometry class:

enter image description here

(The very next proposition, Proposition I.32, is the first time that Euclid uses the parallel postulate - so Euclid was certainly aware that the postulate is not necessary merely to construct parallel lines.)

It's already been pointed out in other answers that Playfair's axiom requires uniqueness of the parallel line. Sometimes Playfair's axiom is even stated in the form "there is at most one parallel line such that", which is equivalent in the presence of the other axioms.


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