七下课本P8-9答案

课本

注意推理步骤模仿

第5题

$解：\because EO\perp AB,\\ \therefore \angle EOB=90^\circ（垂直的定义）.\\ \because \angle EOC=35^\circ,\\ \therefore \angle COB=\angle EOC+\angle EOB\\ =35^\circ+90^\circ=125^\circ.\\ \therefore \angle AOD=\angle COB=125^\circ（对顶角相等）.$

第8题

$解：（1）\because OA平分\angle EOC,\\ \therefore \angle AOC=\frac12 \angle EOC=35^\circ（角平分线的定义）.\\ \therefore \angle BOD=\angle AOC=35^\circ（对顶角相等）.\\ （2）\because \angle EOC+\angle EOD=180^\circ（邻补角的定义）,\\ \angle EOC:\angle EOD=2:3,\\ \therefore \angle EOC=frac25\times 180^\circ=72^\circ,\\ \because OA平分\angle EOC,\\ \therefore \angle AOC=\frac12 \angle EOC=36^\circ（角平分线的定义）.\\ \therefore \angle BOD=\angle AOC=36^\circ（对顶角相等）.$

第13题

解：（1）如图1.
$（2）OE.OF在同一条直线上.\\ 理由如下：\\ \because OE平分\angle AOC,OF平分\angle BOD,\\ \therefore \angle AOE=\frac12 \angle AOC,\\ \angle BOF=\frac12 \angle BOD（角平分线的定义）.\\ \because \angle AOC=\angle BOD,\\ \therefore \angle AOE=\angle BOF,\\ \therefore \angle EOF=\angle AOE+\angle AOD+angle DOF\\ =\angle BOF+\angle AOD+angle DOF\\ =\angle AOB=180^\circ,\\ 即OE.OF在同一条直线上.\\ （3）如图2.~OE\perp OG.\\ 理由如下：\\ \because OG平分\angle AOD,\\ \therefore \angle AOG=\frac12 \angle AOD.\\ 由（1）知, \angle AOE=\frac12 \angle AOC,\\ \therefore \angle EOG=\angle AOE+\angle AOG\\ =\frac12 (\angle AOD+\angle AOC)\\ =\frac12 \angle COD=90^\circ,\\ \therefore ~OE\perp OG.$