An illegal planar Delaunay flip strictly decreases lifted interpolation volume

The two interpolants agree at the four boundary vertices. Their common diagonal intersection $O$ lies strictly inside both diagonals. On the four triangles formed by $O$ and the four boundary edges, each interpolant is affine; their difference is zero at the two boundary endpoints and equals $\delta :=g_{BD}(O)-g_{AC}(O)$ at $O$.

Write $O=(1-t)B+tD$ with $0<t<1$. The plane through $A,B,C$ agrees with $g_{AC}$ at $O$ because $O\in AC$, and its value at $B$ equals the lifted height of $B$. By [A1] its value at $D$ is strictly larger than the lifted height of $D$. Affinity along $BD$ therefore gives $g_{AC}(O)>g_{BD}(O)$, so $\delta <0$.

On a triangle, the integral of an affine function is its area times the average of its three vertex values; this follows by mapping to the reference triangle, where each barycentric coordinate has integral one third of its area. Thus each of the four overlay triangles contributes its area times $\delta \mathrm{/}3$ to the integral of $g_{BD}-g_{AC}$. Their areas sum to the quadrilateral area, so the total difference is $\mathrm{area}(ABCD)\delta \mathrm{/}3<0$.

The row-subtraction and height-column elimination used in [F2] are polynomial determinant identities in the site heights, so they remain valid coefficientwise after each height is replaced by its formal perturbation. Writing ${\mathrm{\Delta }}_D$ for the query height minus that plane's height at $D$ and $O$ for the base orientation determinant, the identity is $OP=-O^2{\mathrm{\Delta }}_D$. The normalized determinant therefore has the opposite sign to ${\mathrm{\Delta }}_D$, so the strict below-plane condition gives a positive, illegal sign. By [F3]–[F4], one common formal configuration resolves a zero constant term. Thus the below-plane condition is exactly the illegal incircle sign and its flip strictly decreases the potential.