Since $\overline{AC} \parallel \overline{DF}$, use alternate interior angles.

$\angle CBF = 34°$ is formed between the transversal $\overline{BF}$ and $\overline{AC}$. By alternate interior angles with the parallel line $\overline{DF}$:

$\angle BFE = 34°$

Since $\triangle EBF$ is isosceles with $BE \cong BF$, the base angles are equal:

$\angle BEF = \angle BFE = 34°$

The angle $\angle BED$ is supplementary to $\angle BEF$ (they form a straight line along $\overline{DE}$):

$\angle BED = 180° - 34° = 146°$