# How to prove that a process ends in a vertex?

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:

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 $$A$$.