VORONOI DIAGRAM BASED DIMENSIONAL ANCHOR ASSESSMENT FOR RADIAL VISUALIZATIONS - Information Visualisation: Techniques, Usability and Evaluation - page 264

Proof of Property 2 : We know the Delaunay triangulation [14] of the
UCBVD must triangulate the convex hull (a convex polygon) of the DAs.
The triangulation's vertices are the d DAs and the barycentre B. Consider
three consecutive adjacent DAs on C (suppose, w.l.o.g., D0, D1, and D2).
As long as D0, D1, and D2 are on C, then the triangle with vertices D0,
D1, D2 can never be in the Delaunay triangulation which is dual to the
UCBVD because C is their circumcircle, and that circumcircle is not
empty since it contains B - a violation of the Delaunay empty circle
property. No triple of non-consecutive DAs can be vertices of the same
triangle in the Delaunay triangulation dual of the UCBVD because B
would be in their circumcircle. The only other possibility for the structure
of a Delaunay triangle is to have two DA vertices together with B. Picking
non-adjacent DAs is not possible, however, since doing so would cause an
edge from an intermediate DA to cross another edge. Edges may not cross
in a triangulation. Each Delaunay triangle has the same structure: vertices
comprised of two adjacent DAs and B. Each triangle's circumcentre is a
Voronoi vertex. Consequently, the line segment connecting each DA to B
is part of the Delaunay triangulation. Each line segment is the dual of an
edge of B's Voronoi region which implies there are d edges to B's
Voronoi region.
Proof of Property 3 : Select any DA and call it D. Draw the bisector b
which separates B from D. We know from Property 2 that some portion of
this bisector must appear as an edge of the Voronoi region sited by B.
Now, consider the arc of the circle which contains D and is determined by
the points where b intersects the circle. Since all DAs are equidistantly
spaced, D is the midpoint of this arc. An easily proven result is that a line
segment drawn from the midpoint of an arc to the centre of the circle is
perpendicular to the chord it subtends. By Property 1, in a UCBVD, B
coincides with the centre of the circle. Thus the segment drawn from B to
any DA is perpendicular to a Voronoi edge.
Proof of Property 4 : Select any DA and call it D, and one adjacent DA
and call it D adj . We may form a quadrilateral by taking the region enclosed
by the segments BD and BD adj and the two Voronoi edges they intersect.
The vertices of this quadrilateral are B, each of the two intersection points
of the Voronoi edges with line segments BD and BD adj , and the
intersection point of the two Voronoi edges. Property 3 informs us that
two interior angles are perpendiculars. Angle DBD adj is known in that it is
the same for any two adjacent DAs since the DAs are placed uniformly on
the circle. Since the other three angles are known, the fourth angle, the
intersection of the two Voronoi edges, may be determined. We find all
interior angles of the BR are the same. Since all angles are the same