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**Lemma. **Given any cyclic polygon, the difference between the number of positively oriented triangles and the number of negatively oriented triangles is independent of the choice of triangulation by diagonals.

**Proof**. First we will show that the result is true for cyclic quadrilaterals (for which there are two choices for the one and only diagonal). For nondegenerate quadrilaterals there are, up to permutation, three configurations to check—the points are arranged in a clockwise, counterclockwise, or criss-cross pattern. In all of these cases the number of positively and the number of negatively oriented triangles does not change, thus neither does the difference. We've illustrated two of the three cases in Figure 13. In Figure 13(a) both triangulations have two negatively oriented triangles and in Figure 13(b) both have one triangle of each orientation.

Figure 13

We must also consider degenerate quadrilaterals—those which have three or fewer vertices.The one and two vertex cases have no nontrivial triangles, so we need only consider the three vertex case. There are four such configurations—two that are triangles and two that are vee-shaped. Two of the cases are shown in Figure 14 (the coincidental points are shown slightly separated for illustration purposes). Figure 14(a) has one degenerate triangle and one positively oriented triangle with both choices of diagonal. Figure 14(b) has a positively and a negatively oriented triangle with one diagonal and two degenerate triangles with the other diagonal. Thus, in these cases, and in the other two that are not pictured, the difference is invariant.

Figure 14

If we have a triangulated cyclic \(n\)-gon and we remove any diagonal, then it leaves an untriangulated quadrilateral. We can then add the other diagonal, which, as we have just shown, does not change the difference in the numbers of positively and negatively oriented triangles. Since it is possible to transform any triangulation into any other triangulation by using this technique repeatedly, the difference is an invariant.∎

**Proof of the Generalized Japanese Theorem.** By the generalized Carnot's theorem for cyclic polygons

\[ \sum_{k=1}^{n-2} \tilde{r}_k = R(q - p) + \sum_{i=1}^n d_i . \]

Clearly \(R\) and the \( d_i \) are independent of the triangulation, and by the previous lemma, so is \( q - p . \) Thus, so is the left-hand side.∎

David Richeson, "The Japanese Theorem for Nonconvex Polygons - A Proof of the Generalized Japanese Theorem," *Loci* (December 2013)