如图,角ACD是三角形ABC的外角,BE平分角ABC,CE平分角ACD,且BE、CE交点于点E,求

2024-11-17 22:24:45
推荐回答(1个)
回答1:

∠E
= 180° - ∠EBC - ∠BCE
= 180° - ∠ABC / 2 - (∠ACB + ∠ACE)

因为∠ACD = 2∠ACE,角平分线,所以
∠E
= 180° - ∠ABC / 2 - ∠ACB - ∠ACD / 2

又因为∠ACD = ∠A + ∠ABC,三角形外角公式,所以
∠E
= 180° - ∠ABC / 2 - ∠ACB - (∠A + ∠ABC) / 2
= 180° - ∠ABC - ∠ABC - ∠A / 2
= (180° - ∠ABC - ∠ABC) - ∠A / 2
= ∠A - ∠A / 2
= ∠A / 2