\because △ACE ≅ △ DCE

\therefore AE = DE, CB⊥AD, CE = BE =\dfrac{1}{2}

According to the Pythagorean Theorem,

AE^2 = AC^2-CE^2

AE =\sqrt{1-\dfrac{1}{4} } = \dfrac{\sqrt{3} }{2}

\therefore AC = 2\times AE = \sqrt{3}