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

2020-06-07 社会 57阅读
∠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
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