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.


 A: Consider it in the complex plane where WLOG,  $C=1,$ $D=e^{2\pi i /3},$ $E=e^{-2 \pi i /3}=\overline D.$  $$(i). \text { We have } \quad 1+D+\overline D=1+D+D^2=0$$  because $0=1-D^3=(1-D)(1+D+D^2)$ and $1-D \ne 0.$
$$(ii). \; \text {  We have }\quad F\overline F=D\overline D=1.$$ Therefore $$FC^2+FD^2+FE^2= |F-C|^2+|F-D|^2+|F-E|^2=$$ $$=(F-C)(\overline F -\overline C)+(F-D)(\overline F-\overline D)+(F-E)(\overline F-\overline E)=$$    $$= (F-1)(\overline F-1)+(F-D)(\overline F-\overline D)+(F-\overline D)(\overline F-D)=$$  $$=6-(F+\overline F)(1+D+\overline D)\quad \text { by } (ii)$$ $$=6 \quad \text { by } (i).$$
A: 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 $$FC^2+FE^2+FD^2=6$$
