How to prove that a process ends in a vertex?

Question

I was playing with regular n-polygons. I noticed this process that always seems to end in one of the vertices:

1. Draw a regular odd-sided polygon with side length $a$.
2. Pick a vertex, $A$, and draw the two largest diagonals going from that point.
3. Choose one of these diagonals, and draw a circle with a radius of $a$ with the center on the other vertex, $B$.
4. Let the "upper" intersection point of this circle and the other diagonal be $C$ and draw a new circle with radius $a$ from that point.
5. Continue 4. until you cannot anymore

It seems that this process terminates always at point $A$ and I am curious why this is. I tried to prove this, but cannot seem to get an elegant approach. I could only prove it for a regular pentagon using trigonometry, but that isn't that nice.

So, why does this end always at point $A$? If you could even provide a more elegant proof for a pentagon or a heptagon, I may be able to continue from that.

Here's the picture for a heptagon I made using GeoGebra:

Answer 1

The large isosceles triangle, with vertex $A$, has vertex angle $\pi/n$, where $n$ is the number of sides of the regular polygon. Angle chasing shows that the $k$-th little isosceles triangle constructed with your process has base angles given by $$ {n+1-2k\over2n}\pi. $$ Hence, after $k=(n-1)/2$ iterations we get a base angle of $\pi/n$, implying that the last triangle has the other base vertex at $A$.