Shame for a TST question. Anyway, $BE/BC = \sqrt 3 - 1$ and $BC/DC = \sqrt 3 \Longrightarrow BE/DC = 3-\sqrt 3$. Taking $DC = 1$ gives $BC=\sqrt 3$ and $BE =3 - \sqrt 3$. Let $AC=x$. $AD=x-1$ and $BC/BE=AC/AE \Rightarrow AE = x(\sqrt 3 -1 )$. $AB/AD = BC/DC = \sqrt 3 \Rightarrow AB = 3-\sqrt 3 + x(\sqrt 3 -1) = (x-1)\sqrt 3 \Rightarrow x = 3$. $AB=(x-1)\sqrt 3 = 2\sqrt 3 \Rightarrow \angle C = 90^\circ, \angle B = 60^\circ, \angle A = 30^\circ$.