证明: (1)如图①，过点$A$作$AG\perp OF$于点$G，$$AH\perp OE$于点$H，$则$\angle AHO=\angle AGO = 90^{\circ}.$$\because \angle EOF = 120^{\circ}，$$\therefore \angle HAG = 60^{\circ}=\angle BAC，$$\therefore \angle HAG-\angle BAH=\angle BAC-\angle BAH，$即$\angle BAG=\angle CAH.$$\because OM$平分$\angle EOF，$$AG\perp OF，$$AH\perp OE，$$\therefore AG = AH.$在$\triangle BAG$和$\triangle CAH$中，$\begin{cases}\angle AGB=\angle AHC \\ AG = AH \\ \angle BAG=\angle CAH\end{cases}，$$\therefore \triangle BAG\cong \triangle CAH(ASA)，$$\therefore AB = AC；$
(2)
(1)中的结论还成立.证明如下：如图②，过点$A$作$AG\perp OF$于点$G，$$AH\perp OE$于点$H，$则$\angle AHC=\angle AGO = 90^{\circ}.$$\because \angle EOF = 120^{\circ}，$$\therefore \angle HAG = 60^{\circ}=\angle BAC.$$\therefore \angle HAG-\angle BAH=\angle BAC-\angle BAH，$即$\angle BAG=\angle CAH.$$\because OM$平分$\angle EOF，$$AG\perp OF，$$AH\perp OE，$$\therefore AG = AH.$在$\triangle BAG$和$\triangle CAH$中，$\begin{cases}\angle AGB=\angle AHC \\ AG = AH \\ \angle BAG=\angle CAH\end{cases}，$$\therefore \triangle BAG\cong \triangle CAH(ASA)，$$\therefore AB = AC；$
(3)①如图③，设点$F，$$M$分别在$BO，$$OA$的延长线上.$\because \angle AOC=\angle BOC = 60^{\circ}，$$\therefore \angle FOA = 180^{\circ}-\angle AOC-\angle BOC = 60^{\circ}，$$\therefore \angle FOA=\angle AOC，$即$OM$平分$\angle COF.$由
(2)知$AC = AB，$$\because \angle BAC = 60^{\circ}，$$\therefore \triangle ABC$是等边三角形；②如图③，在$OC$上截取$ON = OB，$连接$BN.$$\because \angle COB = 60^{\circ}，$$\therefore \triangle BON$是等边三角形，$\therefore BN = OB，$$\angle OBN = 60^{\circ}.$$\because \triangle ABC$是等边三角形，$\therefore \angle ABC = 60^{\circ}=\angle OBN，$$\therefore \angle OBN-\angle ABN=\angle ABC-\angle ABN，$即$\angle ABO=\angle CBN.$在$\triangle AOB$和$\triangle CNB$中，$\begin{cases}BA = BC \\ \angle ABO=\angle CBN \\ BO = BN\end{cases}，$$\therefore \triangle AOB\cong \triangle CNB(SAS)，$$\therefore OA = NC，$$\therefore OC = ON + CN = OB + OA，$即$OC = OA + OB.$
