解：(1)证明：$\because\angle BAC=\angle EAF，$
$\therefore\angle BAC+\angle CAE=\angle EAF+\angle CAE，$即$\angle BAE=\angle CAF。$
在$\triangle BAE$和$\triangle CAF$中，
$\begin{cases}AB = AC \\\angle BAE=\angle CAF \\AE = AF\end{cases}$
$\therefore\triangle BAE\cong\triangle CAF(SAS)。$
$\therefore BE = CF。$
(2)$\because\triangle BAE\cong\triangle CAF，$
$\therefore\angle EBA=\angle FCA，$即$\angle DBA=\angle OCD。$
$\because\angle BDA=\angle ODC，$
$\therefore\angle BAD=\angle COD。$
$\because\angle BAC = 70^{\circ}，$即$\angle BAD = 70^{\circ}，$
$\therefore\angle COD = 70^{\circ}，$即$\angle BOC = 70^{\circ}。$
