10. 如图，直线AB，CD相交于点O，∠AOC = 80°，射线OE把∠BOD分成两个角，且∠BOE∶∠EOD = 3∶5.
（1）求∠BOE的度数；
（2）若过点O作OF⊥OE，求∠BOF的度数.
　　　　

（1）$\because \angle AOC = 80^{\circ}, \angle BOD = \angle AOC, \therefore \angle BOD = 80^{\circ}. \because \angle BOE : \angle EOD = 3 : 5, \therefore \angle BOE = \frac{3}{3 + 5} \times 80^{\circ} = 30^{\circ}$
（2）如图，$\because OF \perp OE, \therefore \angle EOF = 90^{\circ}$. 当$OF$在$\angle AOD$的内部时，$\angle BOF = \angle EOF + \angle BOE = 90^{\circ} + 30^{\circ} = 120^{\circ}$. 当$OF$在$\angle BOC$的内部（即$OF'$处）时，$\angle BOF' = \angle EOF' - \angle BOE = 90^{\circ} - 30^{\circ} = 60^{\circ}$. 综上所述，$\angle BOF$的度数为$60^{\circ}$或$120^{\circ}$

11. 如图，直线AB，CD，EF相交于点O，OG⊥CD，∠BOD = 24°.
（1）求∠AOG的度数.
（2）如果OC是∠AOE的平分线，那么OG是∠AOF的平分线吗？请说明理由.
　　　　

（1）$\because AB, CD$相交于点$O, \therefore \angle AOC = \angle BOD = 24^{\circ}. \because OG \perp CD, \therefore \angle COG = 90^{\circ}$，即$\angle AOC + \angle AOG = 90^{\circ}. \therefore \angle AOG = 90^{\circ} - \angle AOC = 90^{\circ} - 24^{\circ} = 66^{\circ}$ （2）$OG$是$\angle AOF$的平分线 理由：$\because OC$是$\angle AOE$的平分线，$\therefore \angle AOC = \angle COE$. 又$\because \angle DOF = \angle COE, \therefore \angle AOC = \angle DOF. \because OG \perp CD, \therefore \angle COG = \angle DOG = 90^{\circ}. \therefore \angle COG - \angle AOC = \angle DOG - \angle DOF$，即$\angle AOG = \angle FOG. \therefore OG$是$\angle AOF$的平分线.

12. （2024·海门期末）如图，直线CE，DF相交于点P，且CE//OB，DF//OA.
（1）若∠AOB = 45°，求∠PDB的度数.
（2）若∠CPD = 45°，求∠AOB的度数.
（3）将（1）（2）中的∠AOB，∠CPD称为四边形PCOD的一组“对角”，则该四边形的另一组对角相等吗？请说明理由.
　　　　　　

（1）$\because DF // OA, \angle AOB = 45^{\circ}, \therefore \angle PDB = \angle AOB = 45^{\circ}$
（2）$\because CE // OB, \therefore \angle CPD = \angle PDB. \because DF // OA, \therefore \angle PDB = \angle AOB. \therefore \angle AOB = \angle CPD. \because \angle CPD = 45^{\circ}, \therefore \angle AOB = 45^{\circ}$ （3）相等 理由：$\because CE // OB, DF // OA, \therefore \angle OCP + \angle AOB = 180^{\circ}, \angle CPD + \angle ODP = 180^{\circ}. \because \angle AOB = \angle CPD, \therefore \angle OCP = \angle ODP$.

13. （1）如图①，若AB//DE，∠B = 135°，∠D = 145°，求∠BCD的度数.
（2）如图①，在AB//DE的条件下，你能得出∠B，∠BCD，∠D之间的数量关系吗？请说明理由.
（3）如图②，AB//EF，根据（2）中的结论，直接写出∠B + ∠C + ∠D + ∠E的度数.
　　　

（1）如图，过点$C$作$CF // AB$，则$\angle 1 + \angle B = 180^{\circ}. \therefore \angle 1 = 180^{\circ} - \angle B = 180^{\circ} - 135^{\circ} = 45^{\circ}. \because CF // AB, AB // DE, \therefore CF // DE$. 同理，可得$\angle 2 = 180^{\circ} - \angle D = 180^{\circ} - 145^{\circ} = 35^{\circ}. \therefore \angle BCD = \angle 1 + \angle 2 = 45^{\circ} + 35^{\circ} = 80^{\circ}$ （2）$\angle B + \angle BCD + \angle D = 360^{\circ}$ 理由：如图，由（1），知$CF // AB, CF // DE, \therefore \angle B + \angle 1 = 180^{\circ}, \angle D + \angle 2 = 180^{\circ}. \therefore \angle B + \angle 1 + \angle 2 + \angle D = 360^{\circ}$，即$\angle B + \angle BCD + \angle D = 360^{\circ}$. （3）$\angle B + \angle C + \angle D + \angle E = 540^{\circ}$