In $\Delta BEF$,
$\begin{array}{rcll}
BF & = & EF & \text{(given)} \\
\angle EBF & = & \angle BFE & \text{(base $\angle$s, isos. $\Delta$)} \\
\angle EBF & = & \dfrac{1}{2} (180^\circ – \angle BEF) & \text{($\angle$ sum of $\Delta$)} \\
\angle EBF & = & 62^\circ
\end{array}$
Consider $\Delta BCD$.
$\begin{array}{rcll}
BC & = & CD & \text{(property of rhombus)} \\
\angle BCD & = & \angle DBC & \text{(base $\angle$s, isos. $\Delta$)}
\end{array}$
Since $ABCD$ is a rhombus, $AB // DC$ (property of rhombus). Therefore, we have
$\begin{array}{rcll}
\angle ABD & = & \angle BDC & \text{(alt. $\angle$s, $AB // DC$)}
\end{array}$
Since $ABE$ is a straight line, we have
$\begin{array}{rcll}
\angle ABD + \angle CBD + \angle FBE & = & 180^\circ & \text{(adj. $\angle$s on a st. line)} \\
\angle BDC + \angle BDC + 62^\circ & = & 180^\circ & \text{(proved)} \\
\angle BDC & = & 59^\circ
\end{array}$