Since ABCD is a square, $\angle BCD = 90^{\circ}$,

and since CDE is an equilateral triangle, $\angle DCE = 60 ^{\circ}$.

Thus $\angle BCE = \angle BCD + \angle DCE = 90^{\circ}+60^{\circ}=150^{\circ}$.

Because CDE is an equilateral triangle, $EC = DC$ and also, because ABCD is a square, $DC = CB$. Hence $EC = CB$ and ECB is an isosceles triangle.

So $\angle CEB = \angle CBE = \frac{1}{2} (180-150)^{\circ} = 15^{\circ}$, and hence $\angle BED = \angle CED-\angle CEB = 60^{\circ} - 15^{\circ} = 45 ^{\circ}$.

*This problem is taken from the UKMT Mathematical Challenges.*