Thomsen's theorem states that given a triangle ABC, choosing a point on AB (but not A or B) and doing the internal path parallel to AC till reaching BC, and then doing the path parallel to AB till reaching AC, and then doing the path parallel to BC till reaching AB, and then doing the path parallel to AC till reaching BC... and so on, one eventually returns to the initial point.
For visualization, you can check the image (link) from Wikimedia Commons for example:
I have no idea on how to prove this. Maybe it's something very elementary, but I don't see it.
Choose a preferred direction around the triangle, so that a given point separates a side into consistently-ordered "first" and "second" sub-segments. If a given $P_i$ separates a side of the triangle into "first" and "second" sub-segments with lengths in the ratio $1:k$, then $P_{i+1}$ separates the next edge into sub-segments with lengths in the ratio $k:1$.
We have, then, that the ratio flip-flops as the path bounces around a three-sided figure. Every second step of the process returns to the initial ratio, $1:k$, while every third step returns to the initial side. Consequently, the process has "period" $6=2\cdot 3$ (or a divisor thereof), which is to say that points $P_{i}$ and $P_{i+6}$ must coincide.
Note. If $k=1$, then the ratios $1:k$ and $k:1$ match, making the "flip-flops" irrelevant, so that the process has "period" $3$; points $P_i$ and $P_{i+3}$ (and $P_{i+6}$) coincide. (In this special case, we recognize that the $P_i$ are the vertices of the midpoint triangle of $\triangle ABC$.) If $k\neq 1$, then the process has "period" exactly $6$.