Geometric interpretation regarding square of distances. Can anyone give an alternative solution or give a geometric "illustration/interpretation" to the constant relative to the distances(see picture). I could not do it without resorting to coordinates. The constant I found using coordinates is 2*side^2 or 6*radius^2.



Assuming a circle with radius $1$, then the side of the equilateral triangle = $\sqrt3$. By Ptolemy's theorem on the quadrilateral in a circle: $$DE*FC+DC*FE=EC*FD$$ Hence, substituting $DE$ for $DC$ and $EC$, we have$$DE(FC+FE)=DE*FD$$or$$FC+FE=FD$$Squaring, transposing, and factoring gives [1]$$FC^2+FE^2+FD^2=2(FD^2-FC*FE)$$ And since$$EC^2=FC^2+FE^2-2FC*FE*cos\angle EFC$$and $\angle EFC$ is  supplementary to $\angle EDC$, we get$$3=FC^2+FE^2+FC*FE$$or$$FC*FE=3-FC^2-FE^2$$Substituting in [1] above $$FC^2+FE^2+FD^2=2[FD^2-(3-FC^2-FE^2)]$$which leads to $$FC^2+FE^2+FD^2=6$$
