Note that $\angle CBA = \angle ACB = \angle DEC = \angle BFD$, so the triangles $ABC$, $DFB$ and $DCE$ are all similar. We thus have that $FB = BC \times \frac{BD}{AB}$ and $CE = BC \times \frac{DC}{AB}$. Since $AB = AC$, this gives us that $$ BC \times \frac{BD}{AB} + 5AE = AB = AC = AE + BC \times \frac{DC}{AB} $$and so $$ 4AE = BC \times \frac{DC - BD}{AB} = BC \times \frac{BD}{AB}. $$ Thus $$ AB = 5AE + BC \times \frac{BD}{AB} = \frac{9}{4} BC \times \frac{BD}{AB} = \frac{9}{4} \frac{\frac{1}{3} BC^2}{AB} $$and so $$ \left( \frac{AB}{BC} \right)^2 = \frac{3}{4} $$giving us that $$ \frac{AB}{BC} = \frac{\sqrt{3}}{2}. $$