Question

如图，∠ACD是△ABC的外角,BE平分∠ABC,CE平分∠ACD,且BE,CE相交于点E,求证。∠E＝∠A

Answer 1

三角形内角和为180°
∴∠A+∠ABC+∠ACB=180°
∠CBE+∠E+∠BCE=180°
又∵BE平分∠ABC，CE平分∠ACD
∴∠CBE=1/2∠ABC ∠ACE=1/2∠ACD
三角形外角等于不相邻的两内角和，
∴∠ACD=∠A+∠ABC
∠BCE=∠ACB+∠ACE
由∠CBE+∠E+∠BCE=180°可知
1/2∠ABC+∠E+∠ACB+∠ACE=180°
左式=1/2∠ABC+∠E+∠ACB+1/2∠ACD
=1/2∠ABC+∠E+∠ACB+1/2(∠A+∠ABC)
=∠ABC+∠E+∠ACB+1/2∠A
又∵∠A+∠ABC+∠ACB=180°
∴∠A+∠ABC+∠ACB=∠ABC+∠E+∠ACB+1/2∠A
∴∠E=1/2∠A
以上回答你满意么？