That is what I said. But now it seems we are considering Riemannian instead of Euclidean geometry. If you have a Riemannian manifold, the Riemannian metric determines geodesics, two which could, under some circumstances, intersect each other even when they start out “parallel” (as in a line crosses them forming equal angles), like two meridians on a sphere. Note that on a (non-compact) Riemannian manifold you cannot always extend an arbitrary geodesic indefinitely, and even on a geodesically complete manifold things can happen like a geodesic looping back around to intersect itself, so Euclidean geometry no longer applies, except locally.
I am not sure what you mean by this statement, but a Riemannian metric, if we are given one, determines the angle between any two tangent vectors based at the same point. So you know if it measures 90 degrees or not.
Let’s not argue parallel lines, let’s argue about whether .9999… = 1. 
:runs for the hills:
Or philosophy. The simpler the question, the thornier the answer.
Yes; because I say so. And if you disagree you’re a counterrevolutionary traitor who will be shot.
Now, see how easy that was? 
Oh, if we could only enforce that. What a wonderful world it would be.
To me the question makes most sense in the context of Riemannian manifolds as that’s where it makes most sense to talk about “zero curvature”. The example of a cone is just a simple example to show that some degree of caution is needed before assuming the parallel postulate always applies, but also a red herring as the parallel postulate fails at the apex precisely because the Riemannian manifold structure fails at the apex.
Flat (zero curvature) Riemannian manifolds have the property of being locally isometric to the Euclidean n-space, or in other words for every point on such a manifold there exists a neighbourhood that is geometrically equivalent to a neighbourhood of Euclidean n-space. From this a local definition of parallel can be taken from the Euclidean definition of parallel.
I would guess that the parallel postulate may come unstuck when considering flat non-orientable manifold, but I’m waiting for someone with a better grasp of the finer points of geometry and topology to fill in the details. For example a parallel transport of a vector along a flat mobius strip can change it to an antiparallel vector.
If we had a triangle on our mysterious (but known to be flat) surface, and two of the angles were right angles, what could the third angle possibly be? If anything like the Gauss-Bonnet theorem held, it would have to be 0 (or 2π), which seems problematic.
On the subject of analytical algebraic topology of locally Euclidean metrization of infinitely differentiable Riemannian manifolds, wasn’t it proved somewhere that 0.9999… = π which certainly isn’t = 1.
When they converge, Jesus is dragging the stick . . . or something.
CMC fnord!
Yes, but in order to have something like the Gauss-Bonnet theorem apply requires certain conditions.
In fact I can thing of a simple example of a flat manifold where parallel lines intersect:
Take the manifold with boundary defined by taking the set of points in the Euclidean plane {(a,b}), where x[sub]1[/sub]≤a≤x[sub]2[/sub], y[sub]1[/sub]≤b≤y[sub]2[/sub] and (x[sub]2[/sub]-x[sub]1[/sub])>(y[sub]2[/sub]-y[sub]1[/sub]) for some Cartesian coordinate system.
Now make the identification (x[sub]2[/sub],b) = (x[sub]1[/sub]+b,y[sub]2[/sub])
This gives you a flat manifold with boundary where the parallel lines do meet and you can simply remove the boundary to get a flat manifold.
In this example the failure of the parallel postulate can be related to the failure of completeness.
That is essentially your example of before, of a cone, right? Definitely, such conical singularities will modify the application of the Gauss–Bonnet theorem, and one can draw two parallel lines which come back around and intersect at some angle even though the rest of the surface is flat. The OP has not objected to this answer as far as I can interpret the comments.
Note, by the way, that if your space is not a manifold, but, say, an orbifold, then its (appropriately defined) Euler characteristic may not be an integer at all.
If you take the example manifold above and try to complete it (i.e. maximally extend all geodesics) you end up with a conical singularity, but the difference here is that parallel geodesics intersect on the manifold rather than at the singularity.
The example I gave is a flat Riemannian manifold with a Euler characteristic of 0 (it is topologically equivalent to a cylinder), but I believe the Gauss-Bonnet theorem is modified by an additional term for the turning angle of the hole containing the conical singularity.
In a conical space, locally parallel lines never meet at the singularity. They meet somewhere beyond it. And that’s true even if you remove the neighborhood of the singularity from the space.
If you mean that when only a single diffeomorphism to a E[sup]n[/sup] is considered parallel lines don’t meet, then that is true but also trivial as parallel lines don’t meet in E[sup]n[/sup]. Even though that is not the same as saying they can’t meet at singularity as no neighbourhood containing the singularity/a hole can be diffeomorphic to E[sup]n[/sup].
As in, you take a neighborhood of the conical space which doesn’t contain the singularity, and hence is completely flat. In that flat neighborhood, you take two lines which both make the same angle with a line which crosses both of them, and which would therefore be parallel in the infinite flat space to which that neighborhood is homeomorphic. If we then step back again and look at the entire conical space, those two lines will eventually meet, iff the singularity lies between them. But they won’t meet at the singularity.
You can also, of course, have two lines (or rays, at least, since it’s not clear how to extend them past the singularity) which do meet at the singularity. But those lines will nowhere resemble anything that might be called “parallel”.
I think you have to be careful what you mean by “between” the two lines. The singular point could be very far away while you, an ant, walk along a narrow strip defined by two locally parallel lines which eventually intersect back where you started. All the while, you never encountered any singularities along the strip.
I’m still not entirely clear what you mean as two different examples have been given flat Riemannian manifolds with conical singularities
In the first parallel lines don’t meet, though the distance between them goes to zero at the singularity.
In the second the condition for parallel lines to meet is that any neighbourhood containing both lines contains a hole due to the conical singularity, but this hole needn’t be in the strip between the two lines.
Just to be clear: by “lines” I mean line segments.