三、解答题(共40分)
11.(12分)如图,已知直线AB与CD相交于点O,OE,OF分别是$∠ BOD$,$∠ AOD$的平分线.
(1)$∠ DOE$的补角是
$∠AOE$或$∠COE$
;
(2)若$∠ BOD=62°$,求$∠ AOE$和$∠ DOF$的度数.
答案：11.(1)$∠AOE$或$∠COE$
(2)解:$\because OE$是$∠BOD$的平分线,$∠BOD=62^{\circ },$
$\therefore ∠BOE=\frac {1}{2}∠BOD=31^{\circ },$
$\therefore ∠AOE=180^{\circ }-31^{\circ }=149^{\circ }.$
$\because ∠BOD=62^{\circ },\therefore ∠AOD=180^{\circ }-62^{\circ }=118^{\circ }.$
$\because OF$是$∠AOD$的平分线,
$\therefore ∠DOF=\frac {1}{2}∠AOD=\frac {1}{2}×118^{\circ }=59^{\circ }.$
12.(12分)如图,直线AB,CD相交于点O,OC平分$∠ BOE$,$∠ AOE=2∠ FOD$.
(1)若$∠ FOD=21°$,求$∠ AOD$的度数;
(2)猜想OE与OF之间的位置关系,并说明理由.
答案：12.解:(1)$\because ∠FOD=21^{\circ },∠AOE=2∠FOD,$
$\therefore ∠AOE=42^{\circ },$
$\therefore ∠BOE=180^{\circ }-∠AOE=180^{\circ }-42^{\circ }=138^{\circ }.$
$\because OC$平分$∠BOE,$
$\therefore ∠COE=\frac {1}{2}∠BOE=\frac {1}{2}×138^{\circ }=69^{\circ },$
$\therefore ∠AOD=180^{\circ }-∠AOE-∠COE=180^{\circ }-42^{\circ }-69^{\circ }=69^{\circ }.$
(2)$OE⊥OF$.理由如下:
设$∠DOF=x,∠COE=y$,则$∠AOE=2x,∠BOE=2y.$
$\because ∠AOE+∠BOE=180^{\circ },$
$\therefore 2x+2y=180^{\circ },$
$\therefore x+y=90^{\circ }$,即$∠DOF+∠COE=90^{\circ }.$
$\because ∠EOF+∠DOF+∠COE=180^{\circ },$
$\therefore ∠EOF=90^{\circ },\therefore OE⊥OF.$
13.(16分)如图,直线AB,CD相交于点O,OF⊥CD,OE平分$∠ BOD$.
(1)若$∠ AOC=68°$,求$∠ EOF$的度数;
(2)若$∠ AOC=n°(0＜n＜180)$,则$∠ EOF=$
$(90-\frac {1}{2}n)^{\circ }$
;(用含n的代数式表示)
(3)若$∠ BOE$比$∠ BOF$大$24°$,求$∠ COE$的度数.
答案：13.(1)解:$\because ∠AOC=68^{\circ },$
$\therefore ∠BOD=∠AOC=68^{\circ }.$
$\because OE$平分$∠BOD,\therefore ∠DOE=\frac {1}{2}∠BOD=34^{\circ }.$
$\because OF⊥CD,\therefore ∠DOF=90^{\circ },$
$\therefore ∠EOF=∠DOF-∠DOE=90^{\circ }-34^{\circ }=56^{\circ }.$
(2)$(90-\frac {1}{2}n)^{\circ }$
(3)解:设$∠BOF=x^{\circ }$,则$∠BOE=(x+24)^{\circ }.$
$\because OE$平分$∠BOD,$
$\therefore ∠DOE=∠BOE=(x+24)^{\circ }.$
$\because ∠DOF=90^{\circ },$
$\therefore ∠DOE+∠BOE+∠BOF=90^{\circ },$
$\therefore (x+24)+(x+24)+x=90$,解得$x=14,$
$\therefore ∠DOE=38^{\circ },$
$\therefore ∠COE=180^{\circ }-∠DOE=142^{\circ }.$
