解：（1）$\because$ AB = AC，AD $\perp$ BC，$\angle BAC = 120^{\circ}，$
$\therefore$ $\angle BAD = \angle DAC=\frac{1}{2}\angle BAC=\frac{1}{2}\times120^{\circ}=60^{\circ}.$
$\because$ AD = AB，$\therefore$ $\triangle ABD$是等边三角形
（2）$\because$ $\triangle ABD$是等边三角形，
$\therefore$ $\angle ABD = \angle ADB = 60^{\circ}，$BD = AD.
$\because$ $\angle EDF = 60^{\circ}，$$\therefore$ $\angle ADB = \angle EDF.$
$\therefore$ $\angle ADB - \angle ADE = \angle EDF - \angle ADE，$即$\angle BDE = \angle ADF.$
在$\triangle BDE$和$\triangle ADF$中，
$\begin{cases}\angle DBE=\angle DAF = 60^{\circ},\\BD = AD,\\\angle BDE=\angle ADF,\end{cases}$
$\therefore$ $\triangle BDE \cong \triangle ADF.$ $\therefore$ BE = AF
