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第12页

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证明: （1）$\because$ 四边形ABCD为矩形，$\therefore AD// BC，$$AO = CO.$ $\therefore \angle AEO = \angle CFO.$ 在$\triangle AOE$和$\triangle COF$中，$\begin{cases}\angle AEO=\angle CFO,\\\angle AOE=\angle COF,\\AO=CO,\end{cases}$ $\therefore \triangle AOE\cong\triangle COF$ （2）当$\alpha = 90^{\circ}$时，四边形AFCE为菱形 理由：$\because \triangle AOE\cong\triangle COF，$$\therefore OE = OF.$ 又$\because AO = CO，$$\therefore$ 四边形AFCE为平行四边形. 又$\because \angle AOE = 90^{\circ}，$$\therefore AC\perp EF.$ $\therefore$ 四边形AFCE为菱形.

证明: （1）$\because \triangle BOC$绕点C按顺时针方向旋转$60^{\circ}$得到$\triangle ADC，$$\therefore \triangle BCO\cong\triangle ACD，$且$\angle OCD = 60^{\circ}.$ $\therefore OC = DC.$ $\therefore \triangle COD$是等边三角形 （2）$\because \triangle ABC$是等边三角形，$\therefore \angle BAO+\angle OAC = 60^{\circ}，$$\angle ABO + \angle OBC = 60^{\circ}.$ $\because \angle AOB = 105^{\circ}，$$\therefore \angle BAO+\angle ABO = 180^{\circ}-105^{\circ}=75^{\circ}.$ $\therefore \angle OAC+\angle OBC = 60^{\circ}+60^{\circ}-75^{\circ}=45^{\circ}.$ $\because \triangle BCO\cong\triangle ACD，$$\therefore \angle OBC = \angle DAC.$ $\therefore \angle OAD=\angle OAC+\angle DAC=\angle OAC+\angle OBC = 45^{\circ}$ （3）由（1），知$\triangle COD$是等边三角形，$\therefore \angle COD = 60^{\circ}.$ 由（2），知$\angle OAD = 45^{\circ}.$ 当$OA = OD$时，$\angle OAD = \angle ODA = 45^{\circ}，$$\therefore \angle AOD = 90^{\circ}.$ $\therefore \alpha = 360^{\circ}-105^{\circ}-60^{\circ}-90^{\circ}=105^{\circ}.$ 当$OA = AD$时，$\angle AOD=\frac{1}{2}\times(180^{\circ}-45^{\circ}) = 67.5^{\circ}，$$\therefore \alpha = 360^{\circ}-105^{\circ}-60^{\circ}-67.5^{\circ}=127.5^{\circ}.$ 当$OD = AD$时，$\angle AOD = \angle OAD = 45^{\circ}，$$\therefore \alpha = 360^{\circ}-105^{\circ}-60^{\circ}-45^{\circ}=150^{\circ}.$ 综上所述，当$\alpha = 105^{\circ}$或$127.5^{\circ}$或$150^{\circ}$时，$\triangle AOD$是等腰三角形

证明: （1）$\because$ 四边形ABCD是正方形，$\therefore AC\perp BD.$ $\therefore \angle BOE = 90^{\circ}.$ $\because FH\perp AC，$$\therefore \angle EHF = 90^{\circ}=\angle BOE.$ $\therefore \angle BEO+\angle OBE = 90^{\circ}.$ 由旋转得$BE = EF，$$\angle BEF = 90^{\circ}，$$\therefore \angle BEO+\angle HEF = 90^{\circ}.$ $\therefore \angle OBE = \angle HEF.$ 在$\triangle OBE$和$\triangle HEF$中，$\begin{cases}\angle BOE=\angle EHF,\\\angle OBE=\angle HEF,\\BE=EF,\end{cases}$ $\therefore \triangle OBE\cong\triangle HEF$ （2）$\because$ 四边形ABCD是正方形，$AB = 2，$$\therefore$ 易得$OA = \sqrt{2}，$$\angle ACD = 45^{\circ}.$ $\because \triangle OBE\cong\triangle HEF，$$\therefore OE = HF = x.$ $\because \angle FHC = 90^{\circ}，$$\angle HCF = 45^{\circ}，$$\therefore \triangle HCF$是等腰直角三角形. $\therefore HF = CH = x.$ $\therefore CF = \sqrt{HF^{2}+CH^{2}}=\sqrt{2}x.$ $\therefore OE^{2}-CF = x^{2}-\sqrt{2}x=(x - \frac{\sqrt{2}}{2})^{2}-\frac{1}{2}.$ $\because$ 点E在线段AO上（与端点不重合），$OA = \sqrt{2}，$$\therefore 0\lt x\lt\sqrt{2}.$ $\therefore$ 当$x = \frac{\sqrt{2}}{2}$时，$OE^{2}-CF$的最小值是$-\frac{1}{2}$