Question

如图，已知BD为△ABC的角平分线，CD为△ABC的外角∠ACE的平分线，它与BD的延长线交于点D

⑴若∠ABC＝50°,∠ACE＝120°,求：∠D的度数
⑵若∠ACE－∠ABC＝60°,求：∠D的度数
⑶若∠A＝80°求：∠D的度数
⑷猜想：∠A与∠D的数量关系为？

Answer 1

⑴若∠ABC＝50°,∠ACE＝120°,求：∠D的度数
∠D=180°-∠DBC-∠ABC-∠ACD=180°-1/2*50°-(180°-120°)- 1/2*120°=35°
⑵若∠ACE－∠ABC＝60°,求：∠D的度数
∠D=∠ABD+∠BAC-∠ACD
∵∠ACE－∠ABC＝60°, ∠ACE－∠ABC＝∠BAC
∴∠D=∠ABD+∠BAC-∠ACD=60°+1/2∠ABC-1/2∠ACE=60°-1/2(∠ACE－∠ABC)=30°
⑶若∠A＝80°求：∠D的度数∵∠ACE－∠ABC＝∠A＝80°∠D=∠ABD+∠A-∠ACD=80°+1/2∠ABC-1/2∠ACE=80°-1/2(∠ACE－∠ABC)=40°
⑷猜想：∠A与∠D的数量关系为？
∵∠ACE－∠ABC＝∠A
∴∠D=∠ABD+∠A-∠ACD=∠A +1/2∠ABC-1/2∠ACE=∠A -1/2(∠ACE－∠ABC)=
1/2∠A
∠A与∠D的数量关系为∠A=2∠D