如图,直线AB、CD、EF相交于点O,∠AOE:∠AOD=1:3,∠COB:∠DOF=3:4, (1)求∠DOE的度数; (2)试探究CD与EF的关系.

问题描述:

如图,直线AB、CD、EF相交于点O,∠AOE:∠AOD=1:3,∠COB:∠DOF=3:4,

(1)求∠DOE的度数;
(2)试探究CD与EF的关系.

(1)∵∠AOE:∠AOD=1:3,∴设∠AOE=k,∠AOD=3k,则∠COB=∠AOD=3k,∵∠COB:∠DOF=3:4,∴∠DOF=4k,∴∠AOE+∠AOD+∠DOF=k+3k+4k=180°,解得k=22.5°,∴∠DOE=∠AOE+∠AOD=k+3k=4k=4×22.5°=90°,即∠DOE...