Copernicus' Theorem I am trying to understand the proof of Copernicus' Theorem.

Consider two circles of radii R and R/2 with the smaller one rolling
  inside the bigger circle without slipping. Copernicus' Theorem states
  a surprising result that a point on the circumference of the small
  circle traces a straight line segment - a diameter of the big circle,
  to be precise.

If found the following useful resource with a proof.
I was able to understand everything of the proof besides the first argument:

Assume point M on the small circle has previously occupied the
  position of point N on the large circle. Since there is no slipping,
  arcs PM (on the small circle) and PN (on the large circle) have
  exactly the same length.


I can't conclude that the arcs of PM and PN (marked as red) should have the same lenght. Is there a proof for that?

 A: Here's a proof.
Have the point Q be on the opposite side of point N.
Statement: The arcs PN and PM have the same length when the smaller circle rolls without slipping.
Proof:
Assume, to the contrary, that the arcs PN and PM do not have the same length. Then there is no bijection between the points on the arcs PN and PM. As an aide, we can express this (albeit incorrectly) as a function $f: PN→PM$. There are two cases.
Case I: Arc PN has points that correspond to the same point on arc PM.
Case II: Arc PN has all points not correspond to some points on arc PM.
Case I: Let the arc PN have a length greater than or equal to $π$. The smaller circle rolls $2π$ radians or onwards. However, the arc length of PM cannot be greater than the smaller circle’s circumference. From $S=θr$, we have $S=2π \cdot \frac{1}{2}=π$.  Therefore, the point Q is where the arc length is longest. This is precisely the arc length of QN, a contradiction.
Case II: Let the arc PN have a length less than $\pi$. The smaller circle must not touch the larger circle at some point for its arc length to be $\pi$, violating no slipping. ∎
