Solution:

Let $A, B, C, D, E$, and $F$ be six consecutive vertices of the polygon. We prove that, given $A, B, C, D$, and $E$, it is possible to construct $F$ with straightedge alone. Then, continuing around the polygon, we can construct all the vertices and then fill in the sides.

Our proof is based on the observation that the polygon has a line $\ell$ of symmetry such that the pairs $A$ and $F$, $B$ and $E$, $C$ and $D$ are reflections with respect to that line.

The construction proceeds as follows:
- Let $BC$ and $DE$ meet at $X$; let $BD$ and $CE$ meet at $Y$; join $XY$. This is the line $\ell$ of symmetry.
- Let $AB$ meet $XY$ at $U$; join $UE$ (this will pass through $F$, as $AB$ and $EF$ are symmetric about $\ell$).
- Let $AC$ meet $XY$ at $V$; join $VD$ (this likewise passes through $F$).
- The lines $UE$ and $VD$ meet at the point $F$ we seek. (They do not coincide since $D, E, F$ are noncollinear.)