Question

如图,△ABC各角的平分线AD,BE,CF交于点O

Answer 1

（1）证明：∵OB、OC分别平分∠ABC，∠ACB，∴∠OBC=
1
2
∠ABC，∠OCB=
1
2
∠ACB，∴∠BOC=180°-∠OBC-∠OCB=180°-
1
2
（∠ABC+∠ACB）=180°-
1
2
（180°-∠BAC）=180°-90°+
1
2
∠BAC=90°+
1
2
∠BAC；（2）解：∠BOD=∠COG．理由如下：∵△ABC各角的平分线AD，BE，CF交于点O，∴∠ABO=
1
2
∠ABC，∠BAO=
1
2
∠BAC，∠OCG=
1
2
∠ACB，∴∠BOD=∠ABO+∠BAO=
1
2
（∠ABC+∠BAC）=
1
2
（180°-∠ACB）=90°-∠OCG，∵OG⊥BC于G，∴∠OGC=90°，∴∠COG=90°-∠OCG，∴∠BOD=∠COG．

Answer 2

1.∠BOC=180-∠OBC-∠OCB
2∠OBC+2∠OCB=180-∠BAC
∠OBC+∠OCB=90-1/2∠BAC
180-∠OCB-∠OBC=180-(90-1/2∠BAC)=90+1/2∠BAC
2.∠ADB=180-∠ABC-1/2∠BAC
∠BOD=180-∠ADB-1/2∠ABC=1/2∠ABC+1/2∠BAC
∠COG=90-1/2∠ACB
∠ACB=180-∠ABC-∠BAC
∠COG=90-1/2(180-∠ABC-∠BAC)=1/2∠ABC+1/2∠BAC=∠BOD
结论：两角大小相等

Answer 3

(1)∠BOC=180°-∠OBC-OCB=180-180°-1/2(180°-∠BAC)
=180°-90°+1/2∠BAC=90°+1/2∠BAC
(2)∠BOD=∠BAO+∠ABO=1/2(∠ABC+∠BAC)=1/2(180°-∠ACB)
=90°-1/2∠ACB=90°-∠OCD=∠COG

Answer 4

(1)BOD=FAO+FBO=1/2BAC+1/2ABC
COD=OAE+OCE=1/2BAC+1/2ACB
BOC=BAC+1/2(ABC+ACB)=1/2BAC+1/2(BAC+ABC+ACB)=1/2BAC+90
(2)

Answer 5

1.倒角；
2.相等