26. (16分)如图①,点$O在直线AB$上,一把直角三角尺的直角顶点在点$O$处($∠DOE = 90^{\circ}$),三角尺可以绕点$O$转动.
(1)如图②,若$∠BOC = 50^{\circ}$,当三角尺的边$OD落在射线OB$上时,则$∠COE = $______$^{\circ}$；
(2)如图③,若$∠BOC = 120^{\circ}$,$OD始终在∠BOC$的内部,且$∠BOD = 2∠COE$,求$∠AOE$的度数；
(3)若$∠BOC = \alpha(0^{\circ} ＜ \alpha ＜ 180^{\circ}$,且$\alpha ≠ 90^{\circ})$,点$C在直线AB$上方,且直线$OC恰好平分∠AOD$,请画出示意图并直接写出$∠COE$的度数(用含$\alpha$的式子表示).
(1)如图②,若$∠BOC = 50^{\circ}$,当三角尺的边$OD落在射线OB$上时,则$∠COE = $______$^{\circ}$；
(2)如图③,若$∠BOC = 120^{\circ}$,$OD始终在∠BOC$的内部,且$∠BOD = 2∠COE$,求$∠AOE$的度数；
(3)若$∠BOC = \alpha(0^{\circ} ＜ \alpha ＜ 180^{\circ}$,且$\alpha ≠ 90^{\circ})$,点$C在直线AB$上方,且直线$OC恰好平分∠AOD$,请画出示意图并直接写出$∠COE$的度数(用含$\alpha$的式子表示).
答案：
(1)40 (2)如图①，当$OE$在$OC$的右边时，因为$∠BOC = 120^{\circ}$，$OD$始终在$∠BOC$内部，$∠DOE = 90^{\circ}$，所以$∠COE + ∠BOD = 120^{\circ} - 90^{\circ} = 30^{\circ}$。因为$∠BOD = 2∠COE$，所以$∠BOD = 20^{\circ}$。所以$∠AOE = 180^{\circ} - 90^{\circ} - 20^{\circ} = 70^{\circ}$。如图②，当$OE$在$OC$的左边时，因为$∠BOC = 120^{\circ}$，$∠DOE = 90^{\circ}$，所以$∠BOD = 120^{\circ} - ∠COD$，$∠COE = 90^{\circ} - ∠COD$。因为$∠BOD = 2∠COE$，所以$120^{\circ} - ∠COD = 2(90^{\circ} - ∠COD)$。所以$∠COD = 60^{\circ}$。所以$∠COE = 90^{\circ} - 60^{\circ} = 30^{\circ}$。所以$∠AOE = 180^{\circ} - 120^{\circ} - 30^{\circ} = 30^{\circ}$。综上所述，$∠AOE$的度数为$70^{\circ}$或$30^{\circ}$ (3)示意图如图③④所示 如图③，点$K$在直线$OC$上。当$OE$在$OC$的右边时，因为$∠BOC = α$，直线$OC$恰好平分$∠AOD$，所以$∠DOK = ∠AOK = ∠BOC = α$。所以$∠COE = 180^{\circ} - 90^{\circ} - α = 90^{\circ} - α$。如图④，当$OE$在$OC$的左边时，因为$∠BOC = α$，直线$OC$恰好平分$∠AOD$，所以$∠COD = ∠AOC = 180^{\circ} - ∠BOC = 180^{\circ} - α$。所以$∠BOD = 180^{\circ} - 2(180^{\circ} - α) = 2α - 180^{\circ}$。所以$∠COE = ∠BOD + ∠DOE - ∠BOC = 2α - 180^{\circ} + 90^{\circ} - α = α - 90^{\circ}$。综上所述，$∠COE$的度数为$90^{\circ} - α$或$α - 90^{\circ}$
(1)40 (2)如图①，当$OE$在$OC$的右边时，因为$∠BOC = 120^{\circ}$，$OD$始终在$∠BOC$内部，$∠DOE = 90^{\circ}$，所以$∠COE + ∠BOD = 120^{\circ} - 90^{\circ} = 30^{\circ}$。因为$∠BOD = 2∠COE$，所以$∠BOD = 20^{\circ}$。所以$∠AOE = 180^{\circ} - 90^{\circ} - 20^{\circ} = 70^{\circ}$。如图②，当$OE$在$OC$的左边时，因为$∠BOC = 120^{\circ}$，$∠DOE = 90^{\circ}$，所以$∠BOD = 120^{\circ} - ∠COD$，$∠COE = 90^{\circ} - ∠COD$。因为$∠BOD = 2∠COE$，所以$120^{\circ} - ∠COD = 2(90^{\circ} - ∠COD)$。所以$∠COD = 60^{\circ}$。所以$∠COE = 90^{\circ} - 60^{\circ} = 30^{\circ}$。所以$∠AOE = 180^{\circ} - 120^{\circ} - 30^{\circ} = 30^{\circ}$。综上所述，$∠AOE$的度数为$70^{\circ}$或$30^{\circ}$ (3)示意图如图③④所示 如图③，点$K$在直线$OC$上。当$OE$在$OC$的右边时，因为$∠BOC = α$，直线$OC$恰好平分$∠AOD$，所以$∠DOK = ∠AOK = ∠BOC = α$。所以$∠COE = 180^{\circ} - 90^{\circ} - α = 90^{\circ} - α$。如图④，当$OE$在$OC$的左边时，因为$∠BOC = α$，直线$OC$恰好平分$∠AOD$，所以$∠COD = ∠AOC = 180^{\circ} - ∠BOC = 180^{\circ} - α$。所以$∠BOD = 180^{\circ} - 2(180^{\circ} - α) = 2α - 180^{\circ}$。所以$∠COE = ∠BOD + ∠DOE - ∠BOC = 2α - 180^{\circ} + 90^{\circ} - α = α - 90^{\circ}$。综上所述，$∠COE$的度数为$90^{\circ} - α$或$α - 90^{\circ}$