證明: (1)$\because \triangle ADC\cong \triangle BOC，$$\therefore CO = CD.$$\because$將$\triangle BOC$繞點$C$按順時針方向旋轉(zhuǎn)$60^{\circ}$得$\triangle ADC，$$\therefore \angle DCO = 60^{\circ}，$$\therefore \triangle COD$是等邊三角形；

(2)$AD// OC.$理由如下：由

(1)知，$\triangle DOC$是等邊三角形，$\therefore \angle CDO=\angle DOC = 60^{\circ}.$$\because \alpha = 120^{\circ}，$$\triangle COB\cong \triangle CDA，$$\therefore \angle ADC=\angle COB = 120^{\circ}，$$\therefore \angle ADO = 120^{\circ}-60^{\circ}=60^{\circ}，$$\therefore \angle ADO=\angle DOC = 60^{\circ}，$$\therefore AD// OC；$

(3)$\angle AOD = 360^{\circ}-\angle AOB-\alpha-\angle COD = 360^{\circ}-110^{\circ}-\alpha - 60^{\circ}=190^{\circ}-\alpha，$$\angle ADO=\angle ADC-\angle CDO=\alpha - 60^{\circ}，$$\angle OAD = 180^{\circ}-\angle ADO-\angle AOD = 180^{\circ}-(\alpha - 60^{\circ})-(190^{\circ}-\alpha)=50^{\circ}，$若$\angle ADO=\angle AOD，$即$\alpha - 60^{\circ}=190^{\circ}-\alpha，$解得$\alpha = 125^{\circ}；$若$\angle ADO=\angle OAD，$即$\alpha - 60^{\circ}=50^{\circ}，$解得$\alpha = 110^{\circ}；$若$\angle OAD=\angle AOD，$即$50^{\circ}=190^{\circ}-\alpha，$解得$\alpha = 140^{\circ}.$綜上，當(dāng)$\alpha$為$125^{\circ}$或$110^{\circ}$或$140^{\circ}$時，$\triangle AOD$是等腰三角形.

證明: (1)如圖①，過點$A$作$AG\perp OF$于點$G，$$AH\perp OE$于點$H，$則$\angle AHO=\angle AGO = 90^{\circ}.$$\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)中的結(jié)論還成立.證明如下：如圖②，過點$A$作$AG\perp OF$于點$G，$$AH\perp OE$于點$H，$則$\angle AHC=\angle AGO = 90^{\circ}.$$\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如圖③，設(shè)點$F，$$M$分別在$BO，$$OA$的延長線上.$\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$是等邊三角形；②如圖③，在$OC$上截取$ON = OB，$連接$BN.$$\because \angle COB = 60^{\circ}，$$\therefore \triangle BON$是等邊三角形，$\therefore BN = OB，$$\angle OBN = 60^{\circ}.$$\because \triangle ABC$是等邊三角形，$\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.$