Question

如图，在三角形ABC中，角ABC的平分线与角ACB的外角平分线相交于点D，（1）若角A=60度，则角D=多少度

要过程

Answer 1

∵∠A=60°
∴∠ABC+∠ACB=120°
∴∠ACB=120°-∠ABC
∠ACE=∠ABC+∠A=∠ABC+60°
∵∠DBC=1/2∠ABC
∠DCB=∠ACB+1/2∠ACE=(120°-∠ABC)+1/2(∠ABC+60°)
∴∠DBC+∠DCB=1/2∠ABC+(120°-∠ABC)+1/2(∠ABC+60°)=150°
∴∠D=180°-(∠DBC+∠DCB)=180°-150°=30°

Answer 2

30