Divide to n parts

If $S_A,S_B,S_C$ are the sums of the squares of the segments emanating from $A,B,C$ respectively, then $\begin{aligned} S_A & = \; \sum_{j=1}^{n-1} \Big[\left(\frac{ja}{n}\right)^2 + c^2 - \frac{2jac}{n}\cos B\Big] \; =\; \sum_{j=0}^{n-1} \Big[\frac{j^2}{n^2}a^2 + c^2 - \frac{j}{n}(a^2 + c^2 - b^2)\Big] \\ & = \; \sum_{j=1}^{n-1} \Big[\frac{j(j-n)}{n^2}a^2 + \frac{j}{n}b^2 + \frac{n-j}{n}c^2\Big] \end{aligned}$ using the Cosine Rule several times. Similar formulae can be found for $S_B,S_C$ . Thus $S \; = \; S_A + S_B + S_C \; = \; (a^2 + b^2 + c^2)\sum_{j=1}^{n-1}\left(\frac{j^2-nj}{n^2} + \frac{j}{n} + \frac{n-j}{n}\right) \; =\; \frac{(5n-1)(n-1)}{6n}(a^2 + b^2 + c^2)$ and hence $\frac{S}{a^2 + b^2 + c^2} \; = \; \frac{(5n-1)(n-1)}{6n}$ is always rational.