Question

如图,已知AD与BC相交于点o∠ BAD的平分线AE与∠BCD的平分线CE相交于点E,∠B=30°，∠D=50°，求∠AEC

我要第2问！！！谢谢！！！捉急啊！！！！

Answer 1

解：将AE与BC的交点设为F，设∠BAE＝∠1, ∠BCD＝∠2
∵AE平分∠BAD
∴∠BAD＝2∠1
∴∠AOC＝∠B+∠BAD＝∠B+2∠1
∵CE平分∠BCD
∴∠BCD＝2∠2
∴∠AOC＝∠D+∠BCD＝∠D+2∠2
∴∠B+2∠1＝∠D+2∠2
∴2（∠1-∠2）＝∠D-∠B
∵∠BFE＝∠B+∠1, ∠BFE＝∠AEC+∠2
∴∠B+∠1＝∠AEC+∠2
∴∠1-∠2＝∠AEC-∠B
∴2（∠AEC-∠B）＝∠D-∠B
∴∠AEC=（∠B+∠D）/2=(30+40)/2=35度

参考资料http://zhidao.baidu.com/link?url=kXIeOIotxyVeywUt6pAXmbyNgRlH2-3kQEnVsPaw0ofQR5-TBm5Rt4NyhWZVhsFcFR5uuyhJ_IJn2kf3Sg_JpY0813yRibXemcUuXZyVi_K

Answer 2

∠B=30°，∠D=50°，
1/2∠C+∠D=∠AEC+1/2∠A
1/2∠A+∠B=∠AEC+1/2∠C
所以∠D-∠AEC=1/2∠A-1/2∠C=∠AEC-∠B
∠AEC=(∠D+∠B)/2=(30°+50°)/2=40°