Editing Euler's rotation theorem (section)

====Previous analysis====
To arrive at a proof, Euler analyses what the situation would look like if the theorem were true. To that end, suppose the yellow line in '''Figure 1''' goes through the center of the sphere and is the axis of rotation we are looking for, and point {{math|'''O'''}} is one of the two intersection points of that axis with the sphere. Then he considers an arbitrary great circle that does not contain {{math|'''O'''}} (the blue circle), and its image after rotation (the red circle), which is another great circle not containing {{math|'''O'''}}. He labels a point on their intersection as point {{math|'''A'''}}. (If the circles coincide, then {{math|'''A'''}} can be taken as any point on either; otherwise {{math|'''A'''}} is one of the two points of intersection.) 

[[Image:Euler Rotation 2.JPG|200px|right|thumb|'''Figure 2''': Arcs connecting preimage {{math|'''α'''}} and image {{math|'''a'''}} of {{math|'''A'''}} with bisector {{math|'''AO'''}} of the angle at {{math|'''A'''}}.]]

Now {{math|'''A'''}} is on the initial circle (the blue circle), so its image will be on the transported circle (red). He labels that image as point {{math|'''a'''}}. Since {{math|'''A'''}} is also on the transported circle (red), it is the image of another point that was on the initial circle (blue) and he labels that preimage as {{math|'''α'''}} (see '''Figure 2'''). Then he considers the two arcs joining {{math|'''α'''}} and {{math|'''a'''}} to {{math|'''A'''}}. These arcs have the same length because arc {{math|'''αA'''}} is mapped onto arc {{math|'''Aa'''}}. Also, since {{math|'''O'''}} is a fixed point, triangle {{math|'''αOA'''}} is mapped onto triangle {{math|'''AOa'''}}, so these triangles are isosceles, and arc {{math|'''AO'''}} bisects angle {{math|∠'''αAa'''}}.

{{clear}}
[[Image:Euler Rotation 3.JPG|200px|left|thumb|'''Figure 3''': {{math|'''O'''}} goes to {{math|'''O′'''}}, but {{math|'''O′'''}} must coincide with {{math|'''O'''}}.]]