△ABC中,∠A=90°,∠ABC的平分线与∠ACE的平分线相交于点D,求∠D的度数

答案：∠D＝45°
解：因为BD平分∠ABC
故：∠ABD＝∠CBD＝1/2∠ABC
因为CD平分∠ACE
故：∠ACD＝∠DCE＝1/2∠ACE
又：∠DCE＝∠D＋∠CBD；∠ACE＝∠A＋∠ABC
故：∠A＝∠ACE－∠ABC
∠D＝∠DCE－∠CBD＝1/2∠ACE－1/2∠ABC＝1/2(∠ACE－∠ABC)＝1/2∠A
因为∠A=90°
故：∠D＝1/2∠A＝45°

45°

∠D+1/2∠ABC=1/2(90°+∠ABC)

=>∠D+1/2∠ABC=45°+1/2∠ABC
=>∠D=45°

45 就是A的一半