Using the Pythagorean Theorem in $\triangle ABE$ we have that $\overline{AB}^2 + \overline{BE}^2 = \overline{AE}^2$, so $1^2 + (1 - x)^2 = \overline{AE}^2$. Similarly, in $\triangle ECF$ we have $\overline{FC}^2 + \overline{CE}^2 = \overline{EF}^2$, so $x^2 + x^2 = \overline{EF}^2$. Since $\triangle AEF$ is an equilateral triangle, $\overline{AE} = \overline{EF}$, so $1^2 + (1-x)^2 = 2x^2$.