如果∠AOF=15度

设∠AOC=X∵∠COA与∠BOD为对顶角则∠BOD=∠COA=X同理,∠AOD=∠COB=X+45则可得∠COA+∠AOD+∠DOB+∠BOC=360X+X+45+X+X+45=360X=67.5又

前提对角相等所以∠BOC=∠AOD所以∠AOD:∠DOF=3：4因为∠AOE:∠AOD=1:3设∠AOE为X度X+3X+4X=180度X=22.5度所以∠AOF=180-22.5=157.5度

设角1为X6X=180,所以X=30.∠BOD=∠AOC=60因此,∠COE=∠COD-∠1=150.故,∠COF=75,则∠AOF=∠COF+∠AOC=135

∵∠AOC+∠COB=180°∴∠COB=180°-∠AOC∵,∠COB=∠AOC+45°∴180°-∠AOC=∠AOC+45°∴∠AOC=67.5°=∠BOD∴∠COB=112.5°=∠AOD∵OF

∵直线AB∴∠AOB=180∴∠AOC+∠BOC=∠AOB=180∵∠COB=∠AOC+45∴∠AOC+∠AOC+45=180∴∠AOC=135/2∵∠AOC与∠BOD为对顶角∴∠BOD=∠AOC=1

∵直线EF∴∠EOF＝180∴∠EOA+∠AOF＝180∵∠AOF＝4∠EOA∴∠EOA+4∠EOA＝180∴∠EOA＝36∵∠AOC＝90∴∠EOC＝∠AOC-∠EOA＝90-36＝54

/>∵∠AOD：∠BOE=4:1∴设∠AOD=4x,则∠BOE=x∵OE平分∠BOD,∴∠DOE=∠BOE=x,∠BOD=2x∠AOD+∠BOD=180°4x+2x=180°x=30°∴∠BOD=2x

因为角AOD+角BOD=180度,OE平分角BOD,所以角BOE=角BOD/2,因为角AOD：角BOE=4：1,所以角AOD=角BOC=120度,角AOD+角AOC=180度,所以角AOC=60度,因

∠AOC+BOC=180所以∠AOF+∠BOE=90所以∠1+∠2=180-(∠AOF+∠BOE)=90所以∠1和∠2互余∠AOF和∠BOF互补∠AOC和∠BOC互补∠AOE和∠BOE互补∠AOF和∠

解题思路：首先根据角平分线的性质得出∠AOF=∠COF，进而得出∠AOF=∠COF=25°，即可得出答案．解题过程：见附件

∵OE平分∠BOC∴∠COE=∠BOE∵∠DOB∶∠BOE=4∶3∴∠BOD=180°×4÷（4＋3＋3）＝72°∵OF⊥CD∴∠FOD=90°∴∠AOF=180°－∠BOD－∠FOD=180°－72

OE平分∠BOD∠∠BOD=2∠BOE∠AOD:∠BOE=4:1∠AOD:∠BOD=4:2=2:1∠AOD+∠BOD=180°∠AOD=120°∠BOD=60°∠BOE=∠DOE=30°∠AOC=∠B

因为∠AOF+∠FOB=180度,又∠AOF=3∠FOB,所以∠FOB=45度,因为AB⊥CD,所以∠DOB=90度,显然∠AOF=45度,所以对顶角∠EOC=45度

1.解令∠DOE=4x∠BOE=x.则∠DOB=5X所以∠aoc=∠dob=5x（对顶角相等）又因为of平分∠aod因为∠aof=∠dof=180°-5x/2而∠aoc=∠aof-15°所以5x=90

角AOF=135∠AOD:∠BOE=4:1,OE是角平分线,则AOD:BOD=2:1即角AOD=120,AOC=60;BOC=120,BOE=30；则角COE=BOC+BOE=120+30=150,O

45°因为AOC=90°所以AOD=90°因为AOF=3FOBFOB=AOE所以AOE+AOF=4AOE=180°AOE=45°所以EOC=AOC-AOE=90°-45°=45°

∵OE⊥CD于点O，∴∠EOD=90°，∵∠BOE=50°，∴∠BOD=40°，∴∠AOC=40°，∵OD平分角∠BOF，∴∠BOF=80°，∴∠AOF=100°，∴∠EOF=∠EOB+∠BOF=13

121再问：过程啊！！！！%>_

/>∵直线AB∴∠AOB＝180∴∠AOF+∠BOF＝180∵∠AOF＝3∠BOF∴3∠BOF+∠BOF＝180∴∠BOF＝45∵∠BOD与∠AOC为对顶角,∠AOC＝90∴∠BOD＝∠AOC＝90∴

/>∵直线AB∴∠AOB＝180∴∠AOF+∠BOF＝180∵∠AOF＝3∠BOF∴3∠BOF+∠BOF＝180∴∠BOF＝45∵∠BOD与∠AOC为对顶角,∠AOC＝90∴∠BOD＝∠AOC＝90∴