2024-11-15 广告
初一几何证明题.如图
证明:
(1)直接证明:
∵BO平分∠ABC,CO平分∠ACB
∴∠OBC=1/2∠ABC,∠OCB=1/2∠ACB
∴∠BOC
=180°-∠OBC-∠OCB
=180°-1/2∠ABC-1/2∠ACB
=180°-1/2(∠ABC+∠ACB)
=180°-1/2(180°-∠A)
=180°-90°+1/2∠A
=90°+1/2∠A
(2)延长BO交AC于点D
∵∠BOC是△OCD的外角
∴塌嫌∠BOC=∠OCD+∠ODC
∵∠ODC是△ABD的外角
∴∠ODC=∠ABD+∠A
∵BO平分∠ABC,CO平分∠ACB
∴∠ABD=1/2∠ABC,∠OCD=1/2∠ACB
∴∠BOC
=∠OCD+∠ODC
=∠OCD+∠ABD+∠A
=1/2∠ACB+1/2∠ABC+∠A
=1/2(∠ACB+∠ABC)+∠A
=1/2(180°-∠A)+∠A
=90°-1/2∠游衫培A+∠A
=90°+1/2∠A
(3)连结AO并延长与BC交于点E
∵∠BOE是△ABO的外角
∴∠BOE=∠ABO+∠BAO
∵∠神唯COE是△ACO的外角
∴∠COE=∠ACO+∠CAO
∵BO平分∠ABC,CO平分∠ACB
∴∠ABO=1/2∠ABC,∠ACO=1/2∠ACB
∴∠BOC
=∠BOE+∠COE
=∠ABO+∠BAO+∠ACO+∠CAO
=1/2∠ABC+1/2∠ACB+∠BAO+∠CAO
=1/2(∠ABC+∠ACB)+∠A
=1/2(180°-∠A)+∠A
=90°-1/2∠A+∠A
=90°+1/2∠A