How can I improve my explanation that the ratio $\theta=\frac{s}{r}$ that defines the radian measure holds for all circles? I'm trying to demonstrate why the ratio $\theta=\frac{s}{r}$ (where $s$ is an arc measuring some $s$-units in length and $r$ is the radius of the circle) which defines the radian measure holds for all circles.  I'm attempting to show this by using two circles, one of radius $1$ and another radius $r$.
I would like to know if there is anything pertinent that I can add, or that is wrong with this explanation, or perhaps you could put forward a better explanation of your own. I don't think it's rigorous, so go easy. This is what I have so far:


Consider the Unit Circle centered at the origin $O$. Let $P$ be the point on the terminal side of the central angle $\theta$ that lies on the Unit Circle. Let the point $A(1,0)$ be the point on the Unit Circle that lies on the positive $x$-axis.
Consider another circle with radius $r$ (with $r&gt;1$) that is also centered at the origin $O$. Let $Q$ be the point on the terminal side of the central angle $\theta$ that lies on the circle with radius $r$. Let the point $B(r,0)$ be the point on the circle with radius $r$ that lies on the positive $x$-axis.
Here's a diagram:

Given that arcs $QB$ and $PA$ subtend the same central angle $\theta$ they are similar, and as the circle segments are congruent this must mean the radiuses are also similar, so we can write:
$\frac{\text{arc PA}}{\text{arc QB}}=\frac{OP}{OQ}$
As $OP=1$ and $OQ=r$ we have:
$\frac{\text{arc PA}}{\text{arc QB}}=\frac{1}{r}$
Both arc $PA$ and $QB$ are segments of the circumference, and can be written as, $PA=2\pi r k$ where $0&lt;k&lt;1$ and represents the fraction of the circumference that the arc constitutes. As $r=1$ in the case of arc $PA$ we have $PA=2\pi (1) k=2\pi k $. Similarly arc $QB=2\pi r k$, so we now have:
$\frac{2\pi k}{2\pi r k}=\frac{1}{r}$
$\frac{2\pi k}{1}=\frac{2\pi r k}{r}$
This equals the ratio which we have defined to be the radian measure: $\theta=\frac{s}{r}=\frac{2\pi r k}{r}$. Therefore we can conclude.


I lost my way towards the end as I don't know how to conclude this; help would be appreciated. Thanks.
My additions in bold.

Let $\theta$ be any angle. Define the radian measure $\rho(\theta,r)$ to be the ratio $s/r$, where $s$ is the length of the arc which belongs to a circle of radius $r$ and which is bounded by the $x$-axis and a ray through the origin which makes an angle $\theta$ with the $x$-axis (measured in the usual counterclockwise way). We wish to prove that $\rho$ is in fact a constant function of $r$ for any fixed $\theta$ and has therefore the same value for all circles.

Consider the unit circle centered at the origin $O$. Let $P$ be the point on the terminal side of the central angle $\theta$ that lies on the unit circle. Let the point $A(1,0)$ be the point on the unit circle that lies on the positive $x$-axis.

Consider another circle with radius $r$ (with $r\mathbf{\neq} 1$) [your assumption $r&gt;1$ is unnecessary] that is also centered at the origin O. Let $Q$ be the point on the terminal side of the central angle $\theta$ that lies on the circle with radius $r$. Let the point $B(r,0)$ be the point on the circle with radius r that lies on the positive x-axis.

[The rest of your argument.]

This equals the ratio which we have defined to be the radian measure: $\rho(\theta,r)=s/r=2πrk/r=2\pi k=\rho(\theta,1)$. Since $r$ was chosen arbitrarily and $\rho(\theta,1)$ is independent of $r$, we conclude that $\rho(\theta,r)$ is the same for all circles and that we may write $\rho(\theta)$.
Towards the "anything pertinent that I can add" is the use of $\tau = 2\pi$. This reduces the unnecessary repeats of $2\pi$ which makes everything cleaner.