Brazilian Math Olympiad

Let $A_1{A}_2\dots A_{2006}$ be a polygon such that $A_1{A}_{1004}$, $A_2{A}_{1005}$, $\dots$, $A_{1003}{A}_{2006}$ and $M_1{M}_{1004}$, $M_2{M}_{1005}$, $\dots$, $M_{1003}{M}_{2006}$ are concurrent at $O$, where $M_1,M_2,\dots ,M_{2006}$ are the midpoints of $A_1{A}_2,A_2{A}_3,\dots ,A_{2006}{A}_1$, respectively.

Note that $A_1{A}_{1004}$, $A_2{A}_{1005}$, $M_1{M}_{1004}$ are concurrent, so $A_1{A}_2\parallel A_{1004}{A}_{1005}$. Similarly, we obtain that any opposite sides are parallel.

Now we have by applying the ratio theorem repetitively, we obtain $$\frac{O{A}_1}{O{A}_{1004}}=\frac{O{A}_2}{O{A}_{1005}}=\cdots =\frac{O{A}_{1003}}{O{A}_{2006}}=\frac{O{A}_{1004}}{O{A}_1}=\frac{O{A}_{1005}}{O{A}_2}=\cdots =\frac{O{A}_{2006}}{O{A}_{1003}}$$ This implies that $O$ is the midpoint of every main diagonal. Therefore, all opposite sides are congruent and parallel.