Question

如图，直线AB与CD相交于点O，OE垂直AB，OF垂直CD 问：如果（角AOD）=4倍（角EOF），求角EOF的度数。

请写出解答过程

Answer 1

由题意得：
∠AOF+∠EOF=90°
∠EOD+∠EOF=90°

∠EOD+∠DOB=90°

∴∠AOF=∠EOD
∠EOF=∠DOB
∵∠AOD=∠AOF+∠EOF+∠EOD
=∠AOF+∠EOF+∠AOF
=4∠EOF
∴2∠AOF=3∠EOF
∵∠AOF+∠EOF=90°

∴∠EOF=36°

Answer 2

∠AOF+∠EOF=∠AOC+∠AOF=90°
所以∠AOC=∠EOF
所以∠AOC+∠AOD=∠EOF+∠AOD=180
5∠EOF=180
∠EOF=36