如图,在△ABC中,BD平分∠ABC,CD平分△ABC的外角∠ACE,BD、CD相交与点D
问题描述：
如图,在△ABC中,BD平分∠ABC,CD平分△ABC的外角∠ACE,BD、CD相交与点D
（1）∠A=40°时求∠D的度数（2）∠A=x°时,求∠D的度数
答
∠ACE=∠A+∠ABCBD平分∠ABC,CD平分△ABC的外角∠ACE;∠ACD=∠ACE/2=∠A/2+∠ABC/2;∠ACB=180°-∠A-∠ABC;∠D=180°-∠DBC-∠BCD=180°-∠ABC/2-(∠ACB+∠ACD)=180°-∠ABC/2-(180°-∠A-∠ABC+∠A/2+∠ABC/2)=∠A/...能带上因为所以吗。。∵BD平分∠ABC,CD平分△ABC的外角∠ACE;
∴∠ACD=∠ACE/2=∠A/2+∠ABC/2;
∴∠ACB=180°-∠A-∠ABC;
∴∠D=180°-∠DBC-∠BCD
=180°-∠ABC/2-(∠ACB+∠ACD)
=180°-∠ABC/2-(180°-∠A-∠ABC+∠A/2+∠ABC/2)
=∠A/2