TITLE: Proof that Pi is constant (the same for all circles), without using limits
QUESTION [65 upvotes]: Is there a proof that the ratio of a circle's diameter and the circumference is the same for all circles, that doesn't involve some kind of limiting process, e.g. a direct geometrical proof?

REPLY [4 votes]: Let me start by claiming that this is just simply "not true", (while see below if you want a proof.)
In Hyperbolic spaces, the ratio between the circunference and the radius is exponential
In a round Sphere, the ratio between the circunference and the radius is sinusoidal.
So, this means that the ratio between the circunference and the radius is not something that can be easily done by means of simple geometric tools. For instance, the above examples shows that you cannot prove it without the use of the fith Euclid postulate. 
Of course, the proportionality between the circunference and the radius is trivially true if you accept that the procedure of scaling a geometric figure by $\lambda$ scales all the one-dimensional length by $\lambda$.
What you can easily do with all the standard geometric tools is prove it for polygons: For triangles that's just Tale's Intercept Theorem, and a polygon can be easily subdivided in triangles.
Now, if you want to use Thale's theorem for building a propotionality principle for circunferences, than you are forced to introduce limits. 
Note that if you don't want to use limits, then your big problem is to define the lenght of a curve, rather that prove that in the Euclidean space this is scale-multiplicative. 

Finally, if you are more interested in a proof that "hides" limits (for instance for didattic purposes) here is a paper-and-cisor proof of the doubling properties for circunferences:
Consider a disk of paper of radius R. Its circumference has some length L. Now, cut the disk in two halfs: No one has problems in accepting that the two half circunferences have equal lenghts L/2.
If you glue the two radii of the half-disks, you get two identical cones. 
Now you have to convince you audience that if you put one of this cones on the table and look it "from the same level of the table" (i.e. you do a projection) then you see an equilateral triangle!!!
For doing that put your two cones on the table in two different ways: one with the basis on the table, the other with a radius on the table: they will look the same.
This means that the circumference at the base of the cone has radius R/2.
Since we know from the beginning that the base-circumference has length L/2,
We have "proved" that if the circumference of radius R has length L, than the circumference of radius R/2 has lenght L/2. 
By changing the cone angle you get multiplicative constants different from 2 or 1/2, but now convincing your audience will be more tricky.