Let the side of the pentagon be $a$. Then $AC=EC=BD=a\frac{\sqrt{5}+1}{2}.$ The solution is based on using Ptolemy's theorem to cyclic quadrilaterals. From $XABC$: $XA\cdot a+XC\cdot a=XB\cdot a\frac{\sqrt{5}+1}{2}\ (1)$ From $XEDC$: $XE\cdot a+XC\cdot a=XD\cdot a\frac{\sqrt{5}+1}{2}\ (2)$ From $XBCD$: $XB\cdot a+XD\cdot a=XC\cdot a\frac{\sqrt{5}+1}{2}\ (3)$ Adding $(1),(2)$ we get $\frac{\sqrt{5}+1}{2}(XB+XD)=XA+XE+2\cdot XC,$ which (from $(3)$) leads to $XA+XC+XE=\frac{\sqrt{5}+1}{2}(XB+XD)-XC=XB+XC.$ The proof of the general version is identical to the above solution but instead of using Ptolemy's theorem we use Ptolemy's inequality.

Another generalization: Let $n$ be a positive integer. A regular polygon $A_1A_2\dots A_{2n+1}$ is given. The point $X$ is on the circumcircle, on the arc $\stackrel\frown{A_1A_{2n+1}}$. Prove that $\sum_{k=0}^n |XA_{2k+1}|=\sum_{k=1}^n |XA_{2k}|$

math90 wrote: Another generalization: Let $n$ be a positive integer. A regular polygon $A_1A_2\dots A_{2n+1}$ is given. The point $X$ is on the circumcircle, on the arc $\stackrel\frown{A_1A_{2n+1}}$. Prove that $\sum_{k=0}^n |XA_{2k+1}|=\sum_{k=1}^n |XA_{2k}|$ The inequality in this case is also true.

Damarliyaprak wrote: This problem can be solved easily by yahudidolleri theorem. Sincerely,damarliyaprak What is that theorem?

it's a spam I guess

The inequality case and the general case are unsolved here.