In a way, there is no reverse way for you to prove. You must always be careful about what your definitions. That way, you will always have a clear mathematical way of writing what you are trying to prove and how to prove it, as well as how "reverse ways" of some theorems would look.
For example, in your case, you have defined the following:
- regular polygons with $n$ sides.
- The angles of these polygons.
What you have shown is this:
As $n$ approaches infinity, the external angle of a $n$-sided regular polygon approaches $0$.
What you Have not shown is this:
As $n$ approaches infinity, the $n$-sided regular polygon approaches a circle.
Why haven't you shown this? Well, let's see:
First of all, you haven't defined a circle. Sure, you can define a circle, but a circle will not be a regular polygon, at least not by the definition of a regular polygon. OK, that may be a problem you can overcome. We can say that a circle is some sort of curve, just like a polygon. However, there is another problem:
You did not define what it means for two curves to approach one another. Without defining exactly what it means for a polygon to approach a circle, you cannot say a circle approaches it, neither can you then say "How to prove the inverse of this statement?"
Even if you define what it means for two curves to approach each other, you are still miles away from showing the statement which you, ultimately, want to show:
A circle is a regular polygon with an infinite number of sides
Again, why could you not proven this statement? Well, it isn't a mathematical statement, so it cannot be proven in a mathematical way. For example, a polygon is defined as a collection of a finite amount of straight lines, so the concept of "a regular polygon with an infinite number of sides" does not exist yet. You can define it, sure, but if you just define it as a circle, then the statement becomes empty. You could define "generalized polygons" as such:
A curve is a generalized polygon if it is a limit of a sequence of (finite-sided) polygons.
In this case, you must, of course, define what a limit of a curve is, but that is possible (albeit not trivial).
If you decide to define it that way, you can now prove the statement:
If $P_n$ is a regular $n$-sided polygon, then the limit of the sequence $P_1, P_2, \dots$ is a circle.
This statement is, basically, the statement "$n$-sided polygons approach a circle as $n$ tends to infinity", in mathematical terms. However, notice what happened:
- You can no longer speak of a "reverse" of this statement. The statement, by its nature, works only in one direction: the limit of this sequence is this curve.
In defining generalized polygons, you lost the ability to speak about the number of sides of a polygon. For example, a square is a $4$ sided polygon and has $4$ sides. You can say "the number of sides of this polygon is such and such". You cannot say the same thing about generalized polygons. You can define the number of sides as such:
The number of sides of a generalized polygon $P$ is $n$, if $P$ is a $n$-sided polygon, and is $\infty$ if, for all values of $n$, $P$ is not a $n$ sided polygon.
If you decide the number of sides that way, then the statement
A circle has an infinite number of sides
Becomes equivalent to the statement
A circle is a generalized polygon and for all values of $n$, a circle is not a $n$ sided polygon.
Best Answer
A proof depends by the definition of a circle that we use.
If the definition is:
consider a regular polygon with $C$ as center of symmetry. For a point on the polygon the distance $d$ from $C$ is such that: $$ r\le d\le r\cos \dfrac{\theta}{2} $$ where: $r$ is the distance of a vertex form $C$ and $\theta$ is the angle of vertex $C$ subtended by a side.
If the number of sides $n \rightarrow \infty$ than $\theta \rightarrow 0$ and : $$ \lim_{\theta \to 0}r\cos \dfrac{\theta}{2}=r $$ so, at the limit, all points of the polygon have the same distance $r$ from $C$.
if you want use the internal angle $\alpha=\dfrac{\pi}{n}(n-2)$, note that $\theta = \pi -\alpha$ and: $$ n \rightarrow 0 \iff \alpha \rightarrow \pi \iff \theta \rightarrow 0 $$
or use:
$$ r\le d\le r \sin \dfrac {\alpha}{2} $$