$ \pi $, defined as the ratio of a circle's circumference to its diameter, is constant. That is, it does not vary given circles of differing size. This claim is equivalent to saying "all circles are proportional" or "all circles are similar."
Prerequisites
 Laws of similar triangles
 Laws of limits
Proof
Let Circle_{2} represent a circle of radius r_{2}, and construct it. Let Circle_{1} represent a circle of radius r_{1}, such that r_{2} > r_{1} and Circle_{2} is concentric with Circle_{1}, and construct it. Choose an integer $ n \ge 3 $ and let $ \theta = 360^\circ/n $. Construct an isosceles triangle, $ \triangle T_2 $, such that each equal leg has length r_{2}, the vertex adjacent to both equal legs (v_{2}) lies at the center of Circle_{2}, and the measure of the angle at v_{2} is equal to $ \theta $. Let s_{2} represent the length of the side opposite $ \theta $. By this, it can be seen that filling Circle_{2}'s interior with n triangles congruent to $ \triangle T_2 $ such that all share a vertex at the center of Circle_{2}, the equal legs extend outward from that vertex, and none overlap, a regular polygon is created whose perimeter approximates the circumference of Circle_{2}.
By the above constructions, a similar isosceles triangle, $ \triangle T_1 $, is made such that each equal leg has length r_{1}, the vertex adjacent to both equal legs (v_{1}) lies at the center of Circle_{1}, and the measure of the angle at v_{1} is equal to $ \theta $. Let s_{1} represent the length of the side opposite $ \theta $ in $ \triangle T_1 $. Let c_{2} represent the circumference of Circle_{2}, and c_{1} represent the circumference of Circle_{1}. Then:
 $ c_2\approx ns_2 $
 $ c_1\approx ns_1 $
Further, let n increase without bound. Then:
 $ c_2 = \lim_{n \to \infty}ns_2 $
 $ c_1 = \lim_{n \to \infty}ns_1 $
and

(the ratio of Circle_{2}'s circumference to its diameter) 

(the ratio of Circle_{1}'s circumference to its diameter) 
To prove the proposition, it suffices to show that:
 $ {\lim_{n \to \infty}}{ns_2\over 2r_2}=\lim_{n \to \infty}{ns_1\over 2r_1} $
By the laws of similar triangles:
 $ {s_1\over r_1}={s_2\over r_2} $
Substituting:
 $ \lim_{n \to \infty} \frac{ns_2}{2r_2}=\lim_{n \to \infty}\frac{ns_2}{2r_2} $
QED