∠D+∠DBC+∠DCB=180°
∠A+∠ABC+∠ACB=180° (1)
∠DBC=0.5∠ABC
∠DCB=0.5∠ACE+∠ACB=0.5(∠A+∠ABE)+∠ACB=0.5∠A+0.5∠ABE+∠ACB
所以∠D+0.5∠ABC+0.5∠A+0.5∠ABE+∠ACB=180° (2)
(2)式减(1)式得∠D=0.5∠A=40°
∠D+∠DBC+∠DCB=180°
∠A+∠ABC+∠ACB=180° (1)
∠DBC=0.5∠ABC
∠DCB=0.5∠ACE+∠ACB=0.5(∠A+∠ABE)+∠ACB=0.5∠A+0.5∠ABE+∠ACB
所以∠D+0.5∠ABC+0.5∠A+0.5∠ABE+∠ACB=180° (2)
(2)式减(1)式得∠D=0.5∠A=40°