（1）证明：如图①，过点$A$分别作$AG\perp OF$于点$G，$$AH\perp OE$于点$H，$则$\angle AHO = \angle AHC = \angle AGB = \angle AGO = 90^{\circ}。$因为$\angle EOF = 120^{\circ}，$$\angle EOF+\angle AGO+\angle AHO+\angle HAG = 360^{\circ}，$所以$\angle HAG = 60^{\circ}。$又$\angle BAC = 60^{\circ}，$所以$\angle HAG - \angle BAH = \angle BAC - \angle BAH，$即$\angle BAG = \angle CAH。$因为$OM$平分$\angle EOF，$所以$AG = AH。$所以$\triangle BAG\cong\triangle CAH$（ASA）。所以$AB = AC。$
（2）（1）中的结论还成立。证明如下：如图②，过点$A$分别作$AG\perp OF$于点$G，$$AH\perp OE$于点$H，$则$\angle AHC = \angle AHO = \angle AGB = \angle AGO = 90^{\circ}。$因为$\angle EOF = 120^{\circ}，$$\angle EOF + \angle AGO + \angle AHO+\angle HAG = 360^{\circ}，$所以$\angle HAG = 60^{\circ}。$又$\angle BAC = 60^{\circ}，$所以$\angle HAG - \angle BAH = \angle BAC - \angle BAH，$即$\angle BAG = \angle CAH。$因为$OM$平分$\angle EOF，$所以$AG = AH。$所以$\triangle BAG\cong\triangle CAH$（ASA）。所以$AB = AC。$
（3）①证明：如图③，设$F，$$M$两点分别在$BO，$$OA$的延长线上。因为$\angle AOC = \angle BOC = 60^{\circ}，$所以$\angle FOA = 180^{\circ}-\angle AOC - \angle BOC = 60^{\circ}。$所以$\angle FOA = \angle AOC，$即$OM$平分$\angle COF。$由（2），得$AC = AB。$又$\angle BAC = 60^{\circ}，$所以$\triangle ABC$是等边三角形。
②证明：如图③，在$OC$上截取$ON = OB，$连接$BN。$因为$\angle BOC = 60^{\circ}，$所以$\triangle BON$是等边三角形，即$BN = OB，$$\angle OBN = 60^{\circ}。$由（3）①，得$\triangle ABC$是等边三角形，所以$\angle ABC = 60^{\circ}，$$BA = BC。$所以$\angle OBN - \angle ABN = \angle ABC - \angle ABN，$即$\angle ABO = \angle CBN。$所以$\triangle AOB\cong\triangle CNB$（SAS）。所以$OA = NC。$因为$OC = ON + NC，$所以$OC = OA + OB。$
