I don't know how to justify it clearly, but it seems that for the example of equidissection in the wikipedia article you provided, every side of pieces that are on the inside of the square end up outside of the triangle. So it seems an equidissection is at least the perimeter of the triangle, ie. the one provided is optimal.

That's simpler that I expected. It's interesting to notice that it all because the $a_i$ are rationals.

I think that should conclude the question. I was probably not clear about it, but there is no such thing as first and last vertex. Thanks for this simple proof.

@JukkaKohonen You can't do that. You might as well suppose this has to be true wherever you start. There's nothing about step. I don't think it is necessary to conclude

@user44191 indeed !