Let $ABCDEFG$ be a regular heptagon. We'll call the sides $AB$, $BC$, $CD$, $DE$, $EF$, $FG$ and $GA$ opposite to the vertices $E$, $F$, $G$, $A$, $B$, $C$ and $D$ respectively. If $M$ is a point inside the heptagon, we'll say that the line through $M$ and a vertex of the heptagon intersects a side of it (without the vertices) at a $\textit{perfect}$ point, if this side is opposite the vertex. Prove that for every choice of $M$, the number of $\textit{perfect}$ points is always odd.