Hello madyl, this is my insane solution that is probably wrong.
Extend BD to point that we call \(L\) in such that \(BL=BE=EL\), forming equilateral triangle \(BEL\). Copy triangle ABC to the other side of CED to make the entire picture symmetrical.
Since ABC and the other triangle are equilateral and congruent, we know BC = DL. We also know that Angle B = Angle L. Since we made the picture symmetrical, we know that BE = LE. Because of SAS, we know that triangles EBC is congruent to ELD. Because of CPCTC, we know that (B) CE = ED.
kind of weird solution I came up with