The statement clearly holds for the case n = 3 either by the Pascal theorem (parallel lines meet on the line at infinity) or by a simple angle chase. The statement obviously doesn't hold for the case n = 2. As for other values of n, I don't have the answer yet, but I guess one should use induction. Darij

(Sorry I can't use TeX) The statement holds for all odd n>1. Cases n=2 and n=3 are obvious, then it follows by induction. AnAn+1, A2nA1 are parallel iff angle An=angle A2n. It's true iff angle An-1An+2An+1=angle A1A2A2n-1 or it's the same with that An-1An+2, A2n-1A2 are parallel. The statement holds for A1A2...AnAn+1...A2n iff it holds for A2A3...An-1An+2...A2n-1. The statement holds for n iff it holds for n-2.

There is a direct solution as well. We will prove that the statement in question holds if and only if the number n is odd. In fact, we will first show that if n is odd, then the statement holds. In other words, we will show: Lemma 1. If n is an odd positive integer, and a convex 2n-gon $A_1A_2...A_{2n}$ is inscribed into a circle, such that $A_1A_2\parallel A_{n+1}A_{n+2}$, $A_2A_3\parallel A_{n+2}A_{n+3}$, ..., $A_{n-1}A_n\parallel A_{2n-1}A_{2n}$, then $A_nA_{n+1}\parallel A_{2n}A_1$. [Note that hereby, we don't have to exclude the trivial case n = 1: in this case, the assertion $A_nA_{n+1}\parallel A_{2n}A_1$ becomes $A_1A_2\parallel A_2A_1$ and is obviously true.] Proof of Lemma 1. We will work with directed angles modulo 180°. Let X be an arbitrary point on the circumcircle of the 2n-gon $A_1A_2...A_{2n}$. Let g be an arbitrary line in the plane, and for every integer i such that $1\leq i\leq 2n$, let $u_i=\measuredangle\left(A_iA_{i+1};\;g\right)$. Hereby, the indices are cyclic modulo 2n, so that $A_{2n+1}=A_1$ and $u_{2n+1}=u_1$. Note that for every index i, we have $\underbrace{u_i}_{=\measuredangle\left(A_iA_{i+1};\;g\right)}-\underbrace{u_{i+1}}_{=\measuredangle\left(A_{i+1}A_{i+2};\;g\right)} =\measuredangle\left(A_iA_{i+1};\;g\right)-\measuredangle\left(A_{i+1}A_{i+2};\;g\right)$ $=\measuredangle\left(A_iA_{i+1};\;A_{i+1}A_{i+2}\right)=\measuredangle A_iA_{i+1}A_{i+2}$. Since the point X lies on the circumcircle of the 2n-gon $A_1A_2...A_{2n}$, we have $\measuredangle A_iA_{i+1}A_{i+2}=\measuredangle A_iXA_{i+2}$, and thus we can state $u_i-u_{i+1}=\measuredangle A_iXA_{i+2}$. And this must hold for every index i. Hence, $\left(u_1+u_3+...+u_{2n-1}\right)-\left(u_2+u_4+...+u_{2n}\right)$ $=\left(u_1-u_2\right)+\left(u_3-u_4\right)+...+\left(u_{2n-1}-u_{2n}\right)$ $=\measuredangle A_1XA_3+\measuredangle A_3XA_5+...+\measuredangle A_{2n-1}XA_1$ (since $u_i-u_{i+1}=\measuredangle A_iXA_{i+2}$ for every $i$) $=0^{\circ}$. Consequently, $u_1+u_3+...+u_{2n-1}=u_2+u_4+...+u_{2n}$. Since n is odd, this rewrites as $\left(u_1+u_3+...+u_{n-2}+u_n\right)+\left(u_{n+2}+u_{n+4}+...+u_{2n-1}\right)$ $=\left(u_2+u_4+...+u_{n-1}\right)+\left(u_{n+1}+u_{n+3}+...+u_{2n-2}+u_{2n}\right)$. But since $A_1A_2\parallel A_{n+1}A_{n+2}$, we have $\measuredangle\left(A_{n+1}A_{n+2};\;g\right)=\measuredangle\left(A_1A_2;\;g\right)$; in other words, $u_{n+1}=u_1$ (since $u_{n+1} = \measuredangle\left(A_{n+1}A_{n+2};\;g\right)$ and $u_1 = \measuredangle\left(A_1A_2;\;g\right)$). Similarly, $u_{n+2}=u_2$, $u_{n+3}=u_3$, $u_{n+4}=u_4$, ..., $u_{2n-2}=u_{n-2}$, $u_{2n-1}=u_{n-1}$. Hence, $\left(u_1+u_3+...+u_{n-2}+u_n\right)+\left(u_2+u_4+...+u_{n-1}\right)$ $=\left(u_1+u_3+...+u_{n-2}+u_n\right)+\left(u_{n+2}+u_{n+4}+...+u_{2n-1}\right)$ (since $u_2=u_{n+2}$, $u_4=u_{n+4}$, ..., $u_{n-1}=u_{2n-1}$) $=\left(u_2+u_4+...+u_{n-1}\right)+\left(u_{n+1}+u_{n+3}+...+u_{2n-2}+u_{2n}\right)$ $=\left(u_2+u_4+...+u_{n-1}\right)+\left(u_1+u_3+...+u_{n-2}+u_{2n}\right)$ (since $u_{n+1}=u_1$, $u_{n+3}=u_3$, ..., $u_{2n-2}=u_{n-2}$). Most of the terms in this equation cancel out, and what remains is $u_n=u_{2n}$. In other words, $\measuredangle\left(A_nA_{n+1};\;g\right)=\measuredangle\left(A_{2n}A_1;\;g\right)$. Thus, $A_nA_{n+1}\parallel A_{2n}A_1$. Hence, we have proved Lemma 1, i. e. we have proved that the statement is true for n odd. Now, it remains to show that the statement is wrong for n even. This will be done by contradiction: Assume that the statement is true for some even integer n. First note that we can rewrite our statement as follows: Assume we have n + 1 points $A_1$, $A_2$, ..., $A_{n+1}$ on a circle. Let the parallel to $A_1A_2$ through $A_{n+1}$ meet the circle again at $A_{n+2}$. Let the parallel to $A_2A_3$ through $A_{n+2}$ meet the circle again at $A_{n+3}$. Continue constructing new points by the same procedure, until you arrive at the point $A_{2n}$, defined as the second intersection of the parallel to $A_{n-1}A_n$ through $A_{2n-1}$ with our circle. Then, the statement claims that if the 2n-gon $A_1A_2...A_{2n}$ is convex, then $A_nA_{n+1}\parallel A_{2n}A_1$. So, the statement has the form of a polynomial identity (when written out in Cartesian coordinates). Thus, if it is true under the assumption that the 2n-gon $A_1A_2...A_{2n}$ is convex, then it must also be true without this assumption. So it must also be true for self-intersecting or degenerate 2n-gons $A_1A_2...A_{2n}$. But now we can easily construct a degenerate 2n-gon $A_1A_2...A_{2n}$ inscribed into a circle which proves the statement wrong: Let ABCD be a square; let $A_1=A_3=...=A_{n-1}=A$, $A_2=A_4=...=A_n=B$, $A_{n+1}=A_{n+3}=...=A_{2n-1}=D$, $A_{n+2}=A_{n+4}=...=A_{2n}=C$. Then, the conditions $A_1A_2\parallel A_{n+1}A_{n+2}$, $A_2A_3\parallel A_{n+2}A_{n+3}$, ..., $A_{n-1}A_n\parallel A_{2n-1}A_{2n}$ are all satisfied (they are all equivalent to AB || CD, what is trivially true for a square ABCD), but we don't have $A_nA_{n+1}\parallel A_{2n}A_1$ (since this would mean BD || CA, but BD and CA are the diagonals of the square ABCD and thus cannot be parallel). So our statement cannot be true for even n. Hence, we have completely shown that the statement holds if and only if n is odd. The problem is solved. Darij