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?
SKIP TO THE PROOF THAT TWO LINES PERPENDICULAR TO A THIRD ARE PARALLEL TO EACH OTHER TO GET TO THE HEART OF THE PROOF
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.