Question

如图，△ABC的三条内角平分线相交于点O，过点O作OE⊥BC于E点，求证：∠BOD=∠COE

如图，△ABC的三条内角平分线相交于点O，过点O作OE⊥BC于E点，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．