Question

如图直线AB，CD相交于点O,OE平分∠AOD，OF丄CD，垂足为O,∠1+∠2=∠3求∠1，∠2

，∠3？

Answer 1

OE是∠AOD的平分线：∠AOE=∠DOE=∠AOD/2=∠2OF⊥CD：∠COF=∠DOF=90°因为：∠1+∠2=∠3=180°-∠AOD因为：∠1=∠DOF-∠DOB=∠DOF-∠3=90°-∠3=90°-(180°-∠AOD)=∠AOD-90°上两式代入∠1+∠2=∠3有：∠AOD-90°+∠AOD/2=180°-∠AOD所以：(5/2)∠AOD=270°解得：∠AOD=108°所以：∠1=∠AOD-90°=18°∠2=∠AOD/2=54°∠3=180°-∠AOD=72°