(1)探究2结论:∠BOC=
1
2 ∠A,
理由如下:
∵BO和CO分别是∠ABC和∠ACD的角平分线,
∴∠1=
1
2 ∠ABC,∠2=
1
2 ∠ACD,
又∵∠ACD是△ABC的一外角,
∴∠ACD=∠A+∠ABC,
∴∠2=
1
2 (∠A+∠ABC)=
1
2 ∠A+∠1,
∵∠2是△BOC的一外角,
∴∠BOC=∠2-∠1=
1
2 ∠A+∠1-∠1=
1
2 ∠A;
(2)探究3:∠OBC=
1
2 (∠A+∠ACB),∠OCB=
1
2 (∠A+∠ABC),
∠BOC=180°-∠0BC-∠OCB,
=180°-
1
2 (∠A+∠ACB)-
1
2 (∠A+∠ABC),
=180°-
1
2 ∠A-
1
2 (∠A+∠ABC+∠ACB),
结论∠BOC=90°-
1
2 ∠A.
1年前
10