Let $AB = AC = a$ and $BC = \sqrt{2}a$ By trig, we get that $AF = \frac{a}{\sqrt{3}}$, $FC = \frac{(\sqrt{3} - 1)a}{\sqrt{3}}$, $AE = (2 - \sqrt{3})a$ and $BE = (\sqrt{3} - 1)a$ Now, by Ceva's theorem, we get that $\frac{CD}{DB} = 2 - \sqrt{3}$. Since $CD + DC = \sqrt{2}a$, solving, we get that $CD = \frac{(3\sqrt{2} - \sqrt{6})a}{6}$ and $DB = \frac{(3\sqrt{2} + \sqrt{6})a}{6}$ Now, observe that $CD.CB = CF.CA$, which implies that $BDFA$ is cyclic. So, $\angle FDC = \angle FAB = 90^\circ$