Question

在矩形ABCD中,对角线AC BD交与点O,OE平分角ADC交BE于E点,角BDE=15°求角DOC和角COE?

Answer 1

OE平分角ADC -->∠CDE=45

∠CDO=∠CDE-∠BDE=45-15=30

∠DOC=180-∠CDO-∠DCO=180-2*30=120

∠CDO=30 --> CD=2BC,AB=2AD

∠CDE=45 --> AE=AD=AB/2

所以,OE是等腰三角形AOB的中线,

所以,OE是等腰三角形AOB的角平分线 -->∠BOE=∠AOB/2

∠COE=∠COB+∠BOE

=∠COB+∠AOB/2

=∠COB+(180-∠COB)/2

=90+∠COB/2

=90+2∠CDO/2

=90+∠CDO

=90+30

=120