The orientation-normalized lifted determinant records the side of a sphere

Let $A$ be the $d\times d$ matrix with rows $p_i-p_0$ for $i=1,\dots ,d$. Subtracting row $p_0$ from the other rows of the matrix defining $O_d$ leaves its determinant unchanged: the Leibniz formula is linear in each row, and the added determinant has two equal rows, whose terms cancel in pairs under the transposition of those rows. Expansion along the final column then gives $\det A=(-1{)}^d{O}_d$.

Expanding the equal sphere radii in [A1], which exist uniquely by [F3], gives ${\sum }_j{p}_{i,j}^2-{\sum }_j{p}_{0,j}^2=2{\sum }_j{c}_j(p_{i,j}-p_{0,j})$ for $i=1,\dots ,d$. For $q$, put $R={\sum }_j(q_j-c_j{)}^2-r^2$; the same expansion gives ${\sum }_j{q}_j^2-{\sum }_j{p}_{0,j}^2=2{\sum }_j{c}_j(q_j-p_{0,j})+R$.

Subtract row $p_0$ from the other rows of the matrix defining $L_d$, then expand along its final column. The cofactor sign is $(-1{)}^{d+3}$, and the remaining determinant has rows $(p_i-p_0,{\sum }_j{p}_{i,j}^2-{\sum }_j{p}_{0,j}^2)$, including the row for $q$. Subtracting $2c_j$ times coordinate column $j$ from the last column makes its first $d$ entries zero by [step 1.2] and its final entry $R$. The resulting determinant is $\det AR$. These row and column operations preserve the Leibniz determinant: multilinearity in columns follows from the same permutation formula, and the added determinant has equal columns whose terms cancel in pairs under their transposition. Thus $L_d=(-1{)}^{d+3}\det AR=-O_{d}R$.

Multiplication by $O_d$ gives $O_d{L}_d=-O_d^{2}R$. Since $O_d\mathrm{\ne }0$, $O_d^2>0$ in the ordered field [F2]; therefore the product is positive exactly when $R<0$, zero exactly when $R=0$, and negative exactly when $R>0$.