Question

如图7-13,在△ABC中,∠ABC的平分线与∠ACE的平分线交于D点.若∠A=80°,求∠D的度

Answer 1

解：如图，设AC、BD相交于点O
由题可知∠ACE=∠A+∠ABC=80°+∠ABC，∠AOB=∠DOC
∴∠D+∠ACD=∠A+∠ABD,即∠D=80°+∠ABD-∠ACD
∵CD是∠ACE的平分线，BD是∠ABC的平分线
∴∠ACD=∠ACE/2=（80°+∠ABC）/2
∴∠D=80°+∠ABD-∠ACD=80°+∠ABD-（80°+∠ABC）/2=40°+∠ABD-（∠ABC/2）
∵BD是∠ABC的平分线
∴∠ABD=∠ABC/2
∴∠D=40°

Answer 2

：如图，设AC、BD相交于点O
由题可知∠ACE=∠A+∠ABC=80°+∠ABC，∠AOB=∠DOC
∴∠D+∠ACD=∠A+∠ABD,即∠D=80°+∠ABD-∠ACD
∵CD是∠ACE的平分线，BD是∠ABC的平分线
∴∠ACD=∠ACE/2=（80°+∠ABC）/2
∴∠D=80°+∠ABD-∠ACD=80°+∠ABD-（80°+∠ABC）/2=40°+∠ABD-（∠ABC/2）
∵BD是∠ABC的平分线
∴∠ABD=∠ABC/2
∴∠D=40°

Answer 3

解：∵BD平分∠ABC
∴∠ABD=∠DBC
∵DC平分∠ACE
∴∠DCA=∠DCE
∵∠D=180º-∠DBC-∠DCB
=180º-﹙1/2∠ABC+1/2∠ACE+∠ACB﹚
=180º-[1/2∠ABC+2/1(∠A+∠ABC)+∠ACB]
=180º-1/2∠ABC-1/2∠A-1/2∠ABC-∠ACB
=180º-∠ABC-∠ACB-1/2∠A
=1/2∠A
=20º