《2025年名校课堂七年级数学上册北师大版山西专版》

【例】 石家庄外国语校本经典题 如图，$∠AOC=130^{\circ },∠BOC=50^{\circ }$，且 OD 平分$∠AOB$，OE 平分$∠BOC$.
(1)求$∠DOE$的度数.
(2)若将条件“$∠BOC=50^{\circ }$”删去，其余条件不变，则$∠DOE$的度数是否会改变？请说明理由.

【拓展提问】 若 OB 在$∠AOC$的外部，其他条件不变，则$∠DOE=$65°.

答案：
(1)$65^{\circ}$.
(2)不会变.理由如下：$\because OD$平分$\angle AOB,OE$平分$\angle BOC.\therefore \angle DOB = \frac{1}{2}\angle AOB,\angle BOE = \frac{1}{2}\angle BOC.\because \angle AOB + \angle BOC = \angle AOC,\therefore \angle DOE = \angle DOB + \angle BOE = \frac{1}{2}\angle AOB + \frac{1}{2}\angle BOC = \frac{1}{2}\angle AOC = 65^{\circ}$.【拓展提问】 $65^{\circ}$

【变式 1】 如图，已知 OE 平分$∠AOC$，OF 平分$∠BOC$.
(1)若$∠AOB=70^{\circ },∠BOC=30^{\circ }$，则$∠EOF=$35°.
(2)若$∠AOB=\alpha ,∠BOC=\beta $，求$∠EOF$的度数.
(3)通过(1)(2)的计算，能发现什么？

答案：
(1)$35^{\circ}$
(2)$\because OE$平分$\angle AOC,\therefore \angle EOC = \frac{1}{2}\angle AOC.$
$\because OF$平分$\angle BOC,\therefore \angle COF = \frac{1}{2}\angle BOC.\therefore \angle EOF = \angle EOC - \angle COF = \frac{1}{2}\angle AOC - \frac{1}{2}\angle BOC = \frac{1}{2}(\angle AOC - \angle BOC) = \frac{1}{2}\angle AOB = \frac{1}{2}\alpha$.
(3)发现$\angle EOF = \frac{1}{2}\angle AOB.$与$\angle BOC$的度数无关.

【变式 2】 如图，$∠AOB=80^{\circ },∠AOC＜180^{\circ }$，OE 平分$∠AOD$，OF 平分$∠BOC$.若$∠COD=50^{\circ }$，求$∠EOF$的度数.

答案： $\because OE$平分$\angle AOD,OF$平分$\angle BOC,\therefore \angle AOE = \angle EOD = \frac{1}{2}\angle AOD,\angle BOF = \angle FOC = \frac{1}{2}\angle BOC.\therefore \angle EOF = \angle EOD + \angle BOF - \angle BOD = \frac{1}{2}\angle AOD + \frac{1}{2}\angle BOC - \angle BOD = \frac{1}{2}(\angle AOD + \angle BOC) - \angle BOD = \frac{1}{2}(\angle AOB + \angle COD + 2\angle BOD) - \angle BOD = \frac{1}{2}(\angle AOB + \angle COD) = \frac{1}{2} × (80^{\circ} + 50^{\circ}) = 65^{\circ}$.