How can I prove that:

The length of the arc is proportional to the size of the angle.

Every book use this fact in explaining radians and the fundamental arc length equation $s = r\theta$. However no book proofs this fact.

Is this fact some axiom, some natural law like $\pi$ and the triangle side proportions?

Can I proof the above? Or is it something that you should just accept?

  • 1
    $\begingroup$ Euclid proved this in his Elements, Book VI, Proposition 33. $\endgroup$ Jul 15, 2013 at 18:31

7 Answers 7


I believe this is a matter of definition. (One) Radian measure is defined as the angle subtended at the centre by an arc of length equal to the radius of a circle. So for any arc of length $l$, the angle (in radians) is $\theta = \dfrac l r$. So $ l = r\cdot\theta $

  • $\begingroup$ yea..this is the actual proof ..! in fact in the above proof it has not even been considered that , θ is measured in radians ..!and that's a major flaw ..! $\endgroup$
    – arnab
    Nov 23, 2013 at 17:01
  • $\begingroup$ actually θ is the magnitude of the angel when measured in radians ! $\endgroup$
    – arnab
    Nov 23, 2013 at 17:02

Can I prove the above? Or is it something that you should just accept?

Of course you should not "just accept" it. A full formal proof would be a bit more than I'm prepared to give, but a few rather convincing propositions give a good argument.

  1. The arc length is (strictly) monotonic.
  2. The arc length is invariant under rotations.

From the second point, we get

$$l(k\cdot\alpha) = k\cdot l(\alpha)$$

for $k \in \mathbb{N}$ directly, and subdividing, we get

$$l\left(\frac{k}{m}\alpha\right) = \frac{k}{m}l(\alpha)$$

for $k \in \mathbb{N},\; m \in \mathbb{Z}^+$, so the proportionality for rational multiples of the angle $\alpha$

Then, for irrational $t > 0$, we get

$$\sup \{q\cdot l(\alpha)\colon q \in \mathbb{Q}, 0 < q < t\} \leqslant l(t\cdot \alpha) \leqslant \inf\{q\cdot l(\alpha)\colon q \in \mathbb{Q}, q > t\}$$

from the monotonicity.

  • $\begingroup$ One should rather require, in analogy with elementary definitions of length and area, that 1. The length of congruent arcs is the same; 2. The arc length is additive. The monotonicity that you want is then implied. $\endgroup$
    – user35953
    Jul 25, 2020 at 23:47

I am not sure if this is completely true, but I am sure others will correct me if I am mistaken.

Consider a circle of radius R. In polar form, this equation is $r(\theta)=R$. The equation for arc length is as follows:

$L = \int_0^{\theta} \sqrt {r^2+ (\frac {dr} {d\theta})^2 } d\theta$

We know that r=R and dr/d$\theta$ is 0, so the integral becomes:

$L = \int_0^{\theta} R d\theta$


  • $\begingroup$ Interesting. But can you describe how you got the equation for arc length? $\endgroup$ Jul 15, 2013 at 17:42
  • $\begingroup$ tutorial.math.lamar.edu/Classes/CalcII/PolarArcLength.aspx $\endgroup$ Jul 15, 2013 at 17:43
  • $\begingroup$ I like this answer, because thats precisely how I would approach the problem. However, as Parth Tahakkar hints in his comment the problem with this approach is how do you get the equation for arc length? I think that that the equation already presupposes the basic definitions regarding arcs subtended by angles. $\endgroup$
    – Adam
    Nov 18, 2013 at 14:04
  • $\begingroup$ This is fine, but I'd be a little worried that this is circular as the formula for arc length is very likely derived from our assumption that arc length is proportional to angle. $\endgroup$
    – fleablood
    Nov 17, 2015 at 20:45
  • $\begingroup$ So, what formula did the Greeks use before this kind of calculus? $\endgroup$
    – Mark C
    Aug 19, 2018 at 2:10

It should be easy to accept that the length is an additive attribute. That is, if two segments $\ell_1, \ell_2$ are adjacent, then the length of both glued one after the other is $l(\ell_1 \cup \ell_2) = l( \ell_1) + l( \ell_2)$ as long as they do not overlap on a set of measure larger than zero. It is not different for two pieces of arc being next to each other. Then for any union of arcs, one-after-the-other $l (\cup_{i=1}^n s_i)=\sum l(s_i)$, and if they all arcs have the same angle $\theta$, they should have the same length $l(S_{\theta})$(all arcs with the same angle are congruent). So $l( \cup_{i=1}^n s_{\theta_i}) = n l(s_{\theta})$ . Now, for the moment think of an angle $\theta=2 \pi/n$. Then for any angle of the form $\theta=2\pi/n$ we have $n \theta = 2 \pi$, and then: \begin{equation} l( \cup_{i=1}^n S_{\theta,i}) = n l(S_{\theta})= l(S_{2 \pi}) = 2 \pi r, \end{equation} where we accept that the whole circumference is $2 \pi r$, and $\cup_{i=1}^n S_{\theta,i}$ is the union of $n$ congruent arcs into the circle. We then remove the $i$ index after passing to summation since congruent arcs have equal lengths.

From here \begin{equation} l(S_{\theta})= \frac{2 \pi r}{n} = \theta r. \end{equation}

This is good for $\theta$ of the form $2 \pi /n$. Since $l$ is additive then this should be good for any angle of the form $2 \pi p /q$, with $p$ and $q$ in $\mathbb{Z}$ with $q \ne 0$. If $2 \pi/\theta$ is irrational, then this irrational is a limit of a sequence of rational numbers, all of them providing the identical equation $l(S_{\theta}) = \theta r$, so in the limit this equation is valid as well, since the length operator is continuous.

More generally, a function $f$ is linear if:

  1. It is homogenous, $f(s x) = s f(x)$.
  2. It is additive $f(x+y)= f(x)+f(y)$.

It is easy to show that for functions $f:\mathbb{R} \to \mathbb{R}$, homogeneity implies additivity, and so to show linearity it is enough to show homogeneity. In larger spaces (functions of several variables) this is not true. The typical counter-example is a median filter.

The question is; when 2 implies 1, for functions of one variable?

The answer is continuity. If the function is continuous 2 implies 1 and the proof is almost a repetition of the proof above. The interesting and almost unbelieveble fact is that if the continuity hypothesis is removed, the graph of a function $f :\mathbb{R} \to \mathbb{R}$, is dense in $\mathbb{R}^2$. That is, any point of $\mathbb{R}^2$ is as closed as we want to the graph of $f$. This is proved in the following link:

Additive single variable functions

Why is this important on this problem? If the function is linear it is of the from $f(x)=a x$, with some constant of proportionality $a$. Since for the arc function $f(2 \pi)= 2 \pi f(1) =2 \pi r$, then $f(1)=r$, here 1 is units of radians, and BTW, this is the definition or a radian, note that this definition does not account for the proof shown here. Then $f(\theta)=\theta f(1) = r \theta$.

That the arc length function $f=l$ is continuous, can be proved easily from the monotonicity and additivity of the arc length .

It is interesting that this problem is similar to that of the area of a lune (or biangle) in a sphere being directly proportional to $\theta$ with $2 r^2$ being the constant of proportionality and $\theta$ is the dihedral angle of the lune. This problem (of the lune) is usually approached either as a postulate or referred to Euclid's Elements for a proof. Of course when we say Euclid we can not think of formal arguments such as those used above.


These are really two questions.

  1. The length of the arc is proportional to the size of the angle, i.e. $s \propto \theta $ (I)
  2. $ s \propto r $. (II)

(I) is due to the additive property of lengths and angles and two theorems due to Euclid. 1. Equal chords subtend equal angles at the center of a circle. 2. Equal chords of a circle intercepts equal arcs.

These two theorems mean that equal arcs subtend equal angles at the center. The additive property of angles means if we join two equal arcs then the angle subtended by this new arc (of double length) is double the angle subtended by the smaller arcs. A general proportionality $s \propto \theta $ can be proved similarly.

(II) is due to the axiom (I am not sure whether this is an axiom or a theorem, please correct me if I am wrong): The ratio of the circumference to the diameter of a circle is a constant.

This means if the radius is doubled, the circumference is doubled. Now for all other sectors, say semi-circles or sectors subtending an angle $ 60^0 $ at the center, both the smaller and the larger radii sectors' arc-lengths are halved and reduces by a factor 6 respectively. The larger radius sector's arc-length is still double of that of the smaller radius. A general proportionality can be proved similarly.

Note: @Herman Jaramilo gave a correct and more polished answer. Mine may be simpler to follow. Integrals for finding legths is not a valid proof of this result as the length formula in polar form assumes $s = r \theta $ .


It is a linear relationship. $ s = R \theta$. $R$ is the ratio of these two quantities. There is $nothing$ to prove here. It is a $definition$ in flat Euclidean geometry.Definitions are not meant for proving.

In non-euclidean elliptic geometry the arc is less than $ \theta R$ and in the hyperbolic geometry it is more than $ \theta R$, where arcs $s$ are measured on doubly curved surfaces and $\theta$ between principal polar coordinate geodesic lines.

  • 2
    $\begingroup$ @Narasimhan : You can call it a definition but then you should show that is "well defined". That is that there is a proportionality between $s$ and $\theta$ in the circle. How about considring an ellipse instead of a circle? can I say that $s = R \theta$? I can not just say $s=R \theta$ unless I show it makes sense, and this can be formally proved. $\endgroup$ Nov 18, 2015 at 13:39
  • $\begingroup$ @HermanJaramillo Can you provide a link with the proof? $\endgroup$
    – user599310
    Apr 16, 2020 at 19:12

Let o = theta(angle subtended by the arc) Ur equation can be written as s/r=o

We surely have two radii of circle subtending that angle o and forming arc s. In the space between them we can fit more such radii.

The number of that radii will be decided by theta and they will form arc s. Now divide arc s with the radius to get theta.


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