By definition the points on the intersection of Voronoi cells are equidistant from the sites within the cells. The vertex occurs where three or more of these intersections occur. It follows that the vertex is equidistant from the sites of all intersecting Voronoi cells. No site X can be within the circle centered at the vertex with radius r (such that r is the distance between the vertex and the other sites of adjacent Voronoi cells), otherwise the location of the vertex would no longer be on the intersection of the Voronoi cells X and any other site.
I think this could be worded better, but I'm not sure how exactly?
I don't get how to do this?
Let A and B be two points in the intersection, and let point C be on the line segment AB which is not in the intersection (the intersection is not convex).
Then, the point C does not lie within both sets (C is not contained in at least one of the two sets). But, since A and B are in the intersection, they are both within that set lacking C, which contradicts the convexity of that set. This logical contradiction forces us to conclude that the original assumption - that the intersection is nonconvex - is false.
WLOG for three or more intersecting sets.
(So I showed here that "if the intersection is nonconvex then it's false that both sets are convex", which is the contrapositive of "if both sets are convex then so is their intersection".)
I suspect this proof is a lot like the one for 3.11 d.
The boundary of the convex hull (the convex envelope) is made of several connected line segments. Each line segment makes a closed half-plane, and each of these half-planes that intersect at E form the convex hull itself.
That's as far as I get with this one.