如图，△ABC的三条角平分线交于点O，过点O作OE⊥BC于点E，求证：∠BOD=∠COE．

Question

如图，△ABC的三条角平分线交于点O，过点O作OE⊥BC于点E，求证：∠BOD=∠COE．

Answer 1

证明：∵∠AFO=∠FBC+∠ACB=(1)/(2)∠ABC+∠ACB，
∴∠AOF=180°-（∠DAC+∠AF0）
=180°-[(1)/(2)∠BAC+(1)/(2)∠ABC+∠ACB]
=180°-[(1)/(2)（∠BAC+∠ABC）+∠ACB]
=180°-[(1)/(2)（180°-∠ACB）+∠ACB]
=180°-[90°+(1)/(2)∠ACB]
=90°-(1)/(2)∠ACB，
∴∠BOD=∠AOF=90°-(1)/(2)∠ACB，
又∵在直角△OCE中，∠COE=90°-∠OCD=90°-(1)/(2)∠ACB，
∴∠BOD=∠COE．