$FA$ intersect $BC,DE$ at $G,I$ and $BC$ intersect $DE$ at $H$. Then $ABG,CDH,EFI,GHI$ are all equilateral triangles with sides $a,b,c,s$. Then $AB-ED=a-s+b+c=a+b+c-s$ and the rest are the same.

All of the exterior angles of the hexagon are 60 degrees. Then if AF, BC, and DE pairwise intersect (AF-BC, AF-DE, BC-DE) at L, M, and N, respectively, we have equilateral triangles FEN, ABL, CDN, and LMN. Then AB-DE=AB-(MN-ME-DN)=AB-MN+EF+CD. Since there is nothing special about AB and DE, and this equation is cyclic over AB, EF, CD, and we have our desired result.

Problem is an old one: ASU 045 All Russian MO 1964 8.5a 10.3a