Trivial by cosine theorem. Let $ AC = CE = BD = y$ and $ CD = x$. In triangle $ ECD$: $ ED^2 = x^2 + y^2 - 2xy \cos 120 = x^2 + y^2 + xy$. In triangle $ ACB$: $ AB^2 = x^2 + (x + y)^2 - 2x(x + y) \cos 60 = x^2 + x^2 + 2xy + y^2 - x(x + y) = x^2 + x^2 + 2xy + y^2 - x^2 - xy = x^2 + y^2 + xy$. Hence $ AB = DE$ as wanted.

Draw the circle (ABC), name P the middle of the arc AB containing C. Since a 60 degs clockwise rotation about P maps the triangles PCA and PDB, hence PC = PD = CD and <PDB = 120 degs, so the triangles PDB and DCE are equal, hence DE = PB = AB. Best regards, sunken rock