A Voronoi vertex has a common empty circle of nearest sites

Choose three distinct sites $p_1,p_2,p_3$ whose cells contain $x$. From $x\in V(p_i)$, $\mathrm{\parallel }x-p_i{\mathrm{\parallel }}^2\le \mathrm{\parallel }x-p_j{\mathrm{\parallel }}^2$ for every $j$; reversing $i,j$ gives equality. Thus all sites whose cells contain $x$ have the same squared distance $r^2$ from $x$. For every other site $q$, the defining inequality gives $r^2\le \mathrm{\parallel }x-q{\mathrm{\parallel }}^2$, so none is strictly inside the circle.

The common radius is positive, since radius zero would force every site whose cell contains $x$ to equal $x$, contradicting distinctness and the existence of three sites. A line intersects the circle $\mathrm{\parallel }y-x{\mathrm{\parallel }}^2=r^2$ in at most two points: parameterizing the line as $y=a+tv$ gives a nonzero quadratic equation in $t$. Hence no three of the sites on the circle are collinear. For any one such site $p$, the tangent line at $p$ supports the disk and meets the circle only at $p$; therefore $p$ cannot be a convex combination of the other circle sites. All nearest sites are extreme points of their convex hull, so that hull is a polygon with these sites as its vertices.