《2025年1加1轻巧夺冠完美期末七年级数学上册人教版辽宁专版》

20. (8分)(大连西岗期末)
点$O$在直线$AF$上,射线$OB$,$OC$,$OD$,$OE$在直线$AF$的上方,且$\angle AOB = 40^{\circ}$,$\angle AOC = 70^{\circ}$.
(1)如图,$OE$在$\angle COF$的内部,且$OD$平分$\angle BOE$.
①若$\angle COD = 20^{\circ}$,则$\angle EOF =$40°;
②若$\angle COD = n^{\circ}$,则$\angle EOF =$$(80-2n)$$^{\circ}$.(用含$n$的代数式表示)
(2)当$OE$在$\angle BOC$的内部,且$OD$平分$\angle BOE$时,请画出图形;此时,$\angle COD$与$\angle EOF$有怎样的数量关系?请说明理由.

答案：
20.解:
(1)①$40^{\circ}$　[解析]
∵$\angle AOB = 40^{\circ}$，$\angle AOC = 70^{\circ}$，
∴$\angle BOC = \angle AOC - \angle AOB = 70^{\circ} - 40^{\circ} = 30^{\circ}$，
∵$OD$平分$\angle BOE$，$\angle COD = 20^{\circ}$，
∴$\angle DOE = \angle BOD = \angle BOC + \angle COD = 30^{\circ} + 20^{\circ} = 50^{\circ}$，
∵点O在直线AF上，
∴$\angle AOB + \angle BOD + \angle DOE + \angle EOF = 180^{\circ}$，即$40^{\circ} + 50^{\circ} + 50^{\circ} + \angle EOF = 180^{\circ}$，解得$\angle EOF = 40^{\circ}$.
　　②$(80 - 2n)^{\circ}$　[解析]
∵$\angle AOB = 40^{\circ}$，$\angle AOC = 70^{\circ}$，
∴$\angle BOC = \angle AOC - \angle AOB = 70^{\circ} - 40^{\circ} = 30^{\circ}$，
∵$OD$平分$\angle BOE$，$\angle COD = n^{\circ}$，
∴$\angle DOE = \angle BOD = \angle BOC + \angle COD = 30^{\circ} + n^{\circ}$，
∵点O在直线AF上，
∴$\angle AOB + \angle BOD + \angle DOE + \angle EOF = 180^{\circ}$，即$40^{\circ} + 30^{\circ} + n^{\circ} + 30^{\circ} + n^{\circ} + \angle EOF = 180^{\circ}$，
∴$\angle EOF = (80 - 2n)^{\circ}$.
(2)根据题意，作图如答图所示.设$\angle COD = x^{\circ}$，则$\angle BOD = \angle BOC - \angle COD = (30 - x)^{\circ}$，
∵$OD$平分$\angle BOE$，
∴$\angle BOE = 2\angle BOD = (60 - 2x)^{\circ}$，
∴$\angle EOF = \angle BOF - \angle BOE = 140^{\circ} - (60 - 2x)^{\circ} = (80 + 2x)^{\circ}$，即$\angle EOF = 80^{\circ} + 2\angle COD$.