In $\triangle ABC$, $\angle A=55^\circ$, $\angle C=75^\circ, D$ is on side $\overline{AB}$ and $E$ is on side $\overline{BC}$. If $DB=BE$, then $\angle{BED} =$

$\text{(A) } 50^\circ\quad \text{(B) } 55^\circ\quad \text{(C) } 60^\circ\quad \text{(D) } 65^\circ\quad \text{(E) } 70^\circ$
We first consider $\angle CBA$. Because $\angle A = 55$ and $\angle C = 75$, $\angle B = 180 - 55 - 75 = 50$. Then, because $\triangle BED$ is isosceles, we have the equation $2 \angle BED + 50 = 180$. Solving this equation gives us $\angle BED = 65 \rightarrow \fbox{\textbf{(D)}65}$