Looking for elementary proof of "for a circle, $C^2/A = 4 \pi$" If $C$ and $A$ are the circumference and area of a circle, then
$$
\frac{C^2}{A}=4 \pi\; .
$$
I'm looking for a reference to an elementary but rigorous proof of it. 
I'm particularly interested in a proof that is mindful of avoiding, hmm, circularity. :) (Of course I mean the latter in the logical sense, not the geometric.)
This theorem is so well-known that it's a bit difficult to find proofs of it, and it's even harder to find proofs that do not require much more advanced mathematics (e.g., Fourier analysis!) than one typically knows when one first learns it1, or, for that matter, more advanced than what was known at the time the theorem was first established2.

NOTE: a similar expression appears in the so called isoperimetric inequality theorem, which says that 


*

*for any planar figure with perimeter $C$ and area $A$, the following inequality holds:


$$
\frac{C^2}{A} \geq 4 \pi\; ,
$$
...and


*the lower bound is achieved only when the figure is a circle.


This is a different result from the one I'm interested in.

1 I first learned this fact indirectly, in elementary school, in the form of the formulas for the circumference and area of the circle, $C = 2\pi r$ and $A = \pi r^2$ (from which the equation in the title follows easily).  I guess that the same is true for most people.
2 Granted, historically, many theorems of mathematics were routinely invoked before anyone bothered to prove them with any rigor.  Therefore, it would not be surprising if the earliest proofs of this question's theorem relied on mathematics discovered much later than the theorem itself was discovered.
 A: I'm starting with the definition
of $2\pi$ as the total angle
around a point,
or,
equivalently,
as twice the measure
of a straight angle.
As I often do,
I'll play around with an idea
and see where it leads.
Divide a circle
or radius $r$
into $n$ congruent triangles
with one vertex on the center
and two on the circle.
For each of these triangles,
half the angle at the center
is
$\frac{2\pi}{2n}
=\frac{\pi}{n}
$,
so the altitude
of the triangle is
$r\cos(\pi/n)$
and the base is
$2r\sin(\pi/n)$.
The sum of all the bases is
$C_n
=2rn\sin(\pi/n)$.
The area of each triangle
is
$\frac12 \text{base altitude}
=\frac12(2r\sin(\pi/n))(r \cos(\pi/n))
=r^2\sin(\pi/n)\cos(\pi/n)
$
so the total area is
$A_n
=nr^2\sin(\pi/n)\cos(\pi/n)
$.
Therefore
$\dfrac{C_n^2}{A_n}
=\dfrac{4r^2n^2\sin^2(\pi/n)}{nr^2\sin(\pi/n)\cos(\pi/n)}
=\dfrac{4n\sin(\pi/n)}{\cos(\pi/n)}
=4\pi\dfrac{(n/\pi)\sin(\pi/n)}{\cos(\pi/n)}
$.
So,
all that is needed
is to show that
$\lim_{x \to 0} \frac{\sin x}{x}
= 1
$
and
$\lim_{x \to 0} \cos x
= 1
$.
The end.
A: Suppose we already know that $C = 2\pi r$.  Let $A$ be the area of a circle of radius $r$ and let $A+\Delta A$ be the area of a circle of radius $r+\Delta r$, whose circumference is $C+\Delta C= 2\pi(r+\Delta r)$.
Here's a step that I'm not sure how to prove simply and quickly while still being rigorous:
$$
C\,\Delta r \le \Delta A \le (C+\Delta C)\,\Delta r. \tag{basic lemma}
$$
The idea is that $\Delta A$ is the area of a narrow strip of constant width $\Delta r$, a sort of long bent rectangle (a "bent rectangle" is not a rectangle) whose length on one side is $C$ and on the other is $C+\Delta C$, so the area equals the width times something between the two lengths.
Then we have
$$
C\le \frac{\Delta A}{\Delta r} \le C+\Delta C.
$$
Can we show that $\Delta C\to0$ as $\Delta r\to 0$?  That says $C$ is a continuous function of $r$.  That's easy since we know $C=2\pi r$.  Hence by squeezing we get
$$
\frac{dA}{dr} = C.
$$
$$
\frac{dA}{dr} = 2\pi r.
$$
$$
A = \pi r^2 +\text{constant}.
$$
One might find the constant by knowing that $A=0$ when $r=0$, or one might cite a theorem that say the area must be a second-degree homogeneous function of the distances involved.
Next task: How do we prove the basic lemma?
A: Let $A(r)$ be the area of a disk with radius $r$ and $C(r)$ the length of its boundary. Then:
$$ \frac{d}{dr} A(r) = C(r)\tag{1} $$
since the difference between two concentric disks with radii $r,r+\varepsilon$ is a thin annulus with width $\varepsilon$ and length $\approx C(r)$. On the other hand,
$$ A(r) = K r^2 \tag{2} $$
for some constant $K$ that equals the area of the unit disk (i.e. with $r=1$). 
If we call $\color{red}{\pi}$ such a constant, by $(1)$ we have:
$$ A(r) = \color{red}{\pi} r^2,\qquad C(r) = \color{red}{2\pi} r, \tag{3}$$ 
hence:
$$ \frac{C(r)^2}{A(r)} = \color{red}{4\pi} \tag{4}$$
readily follows.
A: As pointed out by David C. Ullrich, much depends on the desired level of rigour in the definitions of area and arc length. For expositions that would be acceptable at school level, see Geometría elemental by Pogorelov, or Geometry by Lang and Murrow, with the first being more rigorous than the second.
Naturally, the level of rigour increases when you look at (rigorous) calculus books, so you might want to have a look at the calculus books by Spivak and Apostol. The main points to look at will be the definitions of arc length and area, as well as of the trigonometric fucntions. Once the appropriate theory is in place, both the area and the arc length of a circle are given by simple integrals.
A: I think the correct statement to prove for a circle of radius $r$, circumference $C$, and area $A$ is:
 $$A = \frac{1}{2} r \, C$$
(otherwise you run into question of the type "... but what is $\pi$?" )
Now you can have your favorite intuitive proof ( using say inscribed and circumscribed polygons). 
I'll try  a "blind" proof using integrals. 
It's enough to show the equality for a circle of radius $1$. One quarter of the area equals the area under the graph of the function $x\mapsto \sqrt{1-x^2}$ for $x\in [0,1]$, that is $\int_0^1\sqrt{1-x^2}\, dx$. The arclength of the parametrization of the circle $x \mapsto (x, \sqrt{1-x^2})$ is $\sqrt{1 + \left(-\frac{x}{\sqrt{1-x^2}}\right)^2}= \frac{1}{\sqrt{1-x^2}}$. Hence we have to show the equality:
$$\int_0^1\sqrt{1-x^2}\, dx=\frac{1}{2} \int_0^1 \frac{dx}{\sqrt{1-x^2}}\, dx$$ 
Using integration by parts, LHS 
$$\int_0^1\sqrt{1-x^2}\, dx= x\sqrt{1-x^2}\mid_0^1 + \int_0^1\frac{x^2}{\sqrt{1-x^2}}= \int_0^1\frac{x^2}{\sqrt{1-x^2}}$$
However, LHS also equals
$$\int_0^1\sqrt{1-x^2}\, dx=\int_{0}^1\frac{1-x^2}{\sqrt{1-x^2}} \, dx$$
hence the equality.
