Rotations and the parallel postulate. If we take a full rotation to be $360^\circ$, then it seems that we can prove the following

Starting from the red point, we walk clockwise along the triangle. At each vertex, we must turn through the green angles marked to proceed down the adjacent sides of the triangle. When we return to the red point, we will have turned through one full rotation. This means that the sum of the exterior angles is given as $360^\circ$, implying the interior angles of the triangle sums of $180^\circ$.
The fact that the angles of a triangle sum to $180^\circ$ is well known to be equivalent to the parallel postulate and this made me wonder whether if the fact that a full rotation being $360^\circ$ is also equivalent to the parallel postulate?
I avoided stating the question using "exterior angles of a triangle sums to $360^\circ$" and instead used the more ambiguous term "rotations" to emphasize the fact that rotations seem to be more general. We can for example show that the interior angles of a heptagram sum to $180^\circ$ by noting that three full rotations are made while "walking" thge heptagram. This should generalize to arbitrary closed polygons and seems stronger than the fact that the exterior angles sum to $180^\circ$.
In summary, I would be interested in knowing the connections that this technique has to the parallel postulate as well as if this technique is a "rigorous" way of finding the internal angles of more complex shapes such as the heptagram.
 A: Your picture, and perhaps your assumptions, are lying in the Euclidean plane. Take the same idea and put it on the sphere, where the parallel postulate is false, and we get something like the following:

Notice that, in this case, the sum of the exterior angles is $270^\circ$, not $360^\circ$.
However, in answer to your question about the sum of the interior angles of a polygon, since an external and the corresponding interior angle sum to $180^\circ$, the sum of the exterior angles and the interior angles is $180^\circ\times$ the number of sides. Since, as you have noted, in the Euclidean plane, the sum of the exterior angles is $360^\circ$, we get that the sum of the interior angles of a polygon with $n$ sides is $(n-2)180^\circ$.
A: Good question. To begin with, the first part of the question is false. The "fact" that a full rotation is 360 degrees, is in fact a definition of the angular measure of a full rotation, and thus is not equivalent to the parallel postulate. You can think of it as defining what a degree is, namely, one part in 360 of a full rotation. We could have defined it to be say 100 "new degrees", but as it was, the ancients defined it as 360 degrees, presumably as a good approximation to one rotation of the earth around the sun. Thus one degree of rotation corresponds to (roughly) one day along earth's annual orbit around the sun.
For the last part of your question, indeed, this technique can be used to prove the connection between the parallel postulate and the interior angles sum of a triangle (and by extension to other polygons etc). The point is that to make your "walk clockwise along the triangle...the sum of the exterior angles is given as  360∘" statement rigorous, you need to use the parallel postulate to put all these exterior angles to the same point (ie to parallel transport all the exterior angles to one local point, and see that all the exterior angles add up to one full rotation), just that this idea is now so innate that you had implicitly assumed its validity, and as the other answers have shown, this is only true on the Euclidean plane.
