如图所示,已知：三角形ABC中,BD平分角ABC,CD平分三角形ABC的外角角ACE,BD、CD交于点D

1、∵1/2∠ACE=∠D+1/2∠ABC∠ACE=∠A+∠ABC∴1/2(∠A+∠ABC)=∠D+1/2∠ABC1/2∠A+1/2∠ABC=∠D+1/2∠ABC∴∠D=1/2∠A 2、∵AB∥CD∴∠DCE=1/2∠ACE=∠ABC∵∠ACE=∠A+∠ABC∴1/2(∠A+∠ABC)=∠ABC1/2∠A+1/2∠ABC=...还有一问回答了