Question

如图,在△ABC中,BD平分∠ABC,CD平分△ABC的外角∠ACE,BD、CD相交与点D

如图,在△ABC中,BD平分∠ABC,CD平分△ABC的外角∠ACE,BD、CD相交于点D
（1）当∠A=40°，90°，130°时，请你分别求出∠D的度数。（写出详细的解题过程）
（2）根据以上求解过程，你能发现∠A与∠D的大小之间存在什么规律么？写出你的发现

Answer 1

∠ACE=∠A+∠ABC
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
(1)∠A=40°，∠D=∠A/2=20°；
∠A=90°，∠D=∠A/2=45°；
∠A=130°，∠D=∠A/2=65°；
(2)∠D=∠A/2.

Answer 2

∠ACE=∠A+∠ABC
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
(1)∠A=40°，∠D=∠A/2=20°；
∠A=90°，∠D=∠A/2=45°；
∠A=130°，∠D=∠A/2=65°；
(2)∠D=∠A/2.

Answer 3

(1) 20° , 45° , 65°
(2) ∠D=∠DCE-∠DBC=0.5∠ACE - 0.5∠ABC=0.5*∠A

Answer 4

∠ACE=∠A+∠ABC
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
(1)∠A=40°，∠D=∠A/2=20°；
∠A=90°，∠D=∠A/2=45°；
∠A=130°，∠D=∠A/2=65°；
(2)∠D=∠A/2.

Answer 5

∠ACE=∠A+∠ABC
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
(1)∠A=40°，∠D=∠A/2=20°；
∠A=90°，∠D=∠A/2=45°；
∠A=130°，∠D=∠A/2=65°；
(2)∠D=∠A/2.