Let the trapezoid be $ABCD$ with $AB = 1 = BC = AD = 1, CD = AC = BD = a$. Then let $\angle DAB = \alpha$, so $$\angle DBC = \alpha - (180^{\circ} - \alpha)/2 = \angle DCB = 180^{\circ} - \alpha \implies \alpha = 108^{\circ}.$$Then $a = 2 \sin 54^{\circ} = \boxed{\frac{\sqrt{5} + 1}{2}}$.

Label the trapezoid the same as the above solution. Since the trapezoid is isosceles, it is cyclic. Then, by Ptolemy's theorem, we have $a^2=a+1 \implies a=\frac{1 \pm \sqrt{5}}{2}.$ We discard the negative value, so $a=\boxed{\frac{1 + \sqrt{5}}{2}}.$