Question

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

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，求证：角E等于二分之一角A

Answer 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