Question

在三角形ABC中,角B的平分线与角C的外角平分线交于D,且角D=30度

在三角形ABC中,角B的平分线与角C的外角平分线交于D,且角D=30度,求角A的度数.!

急!

Answer 1

∠DCE=∠DBE+∠D
=∠ABD+∠D
=∠ACD

又∠A+∠ABD=∠D+∠ACD
=∠D+∠ABD+∠D
=2∠D+∠ABD

∴∠A=2∠D=60°

Answer 2

∠ABD=∠DBE,∠ACD=∠DCE
∠DBE+∠ACB+∠ACD=180-∠D=150
∠DBE+∠ACB+(180-∠ACB)/2=150
∠DBE+1/2∠ACB=60
1/2∠ABC+1/2∠ACB=60
∠ABC+∠ACB=120
∠A=60