Question

BD、CE分别是△ABC中∠ABC和∠ACB的平分线，且BD、CE相交于点O，∠BAC=80°，求∠BOC的度数

要有完整整过程和条理性！

Answer 1

∠BOC=130°
∠BOC=∠BDC+∠ACE
∠BDC=∠A+∠ABD
∠ACE=1/2∠ACB
∠BOC=∠A+∠ABD+1/2∠ACB
∠ABD=1/2∠ABC
∠BOC=∠A+1/2∠ABC+1/2∠ACB
∠BOC=1/2∠A+1/2∠A+1/2∠ABC+1/2∠ACB
∠BOC=1/2(∠A+∠ABC+∠ACB)+1/2∠A=90° +1/2∠A=90° +40° =130°

Answer 2

连续应用三角形外角定理及角平分线定义。
∠BOC=∠ODC+∠ACE(三角形的外角等于与它不相邻的两个内角和)
=(∠A+∠ABD)+∠ACE(同上)
=∠A+1/2∠ABC+1/2∠ACB(角平分线定义 )
=∠A+1/2(∠ABC+∠ACB)(提取1/2)
=∠A+1/2(180°-∠A)(三角形的内角和等于180°)
=90°+1/2∠A(去括号，合并)
=90°+40°
=130°。

Answer 3

∠ABC+∠ACB=180°-80°=100°
(∠ABC+∠ACB)÷2=50° (BD,CE是角平分线)
∠BOC=180°-50°=130°

Answer 4

100

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