Question

BD,CD分别是△ABC的两个外角∠CBE,∠BCF的平分线是探求∠BDC与∠A之间的数量关系

Answer 1

∵∠CBE=∠A+∠ACB ∠BCF=∠A+∠ABC ∠A+∠ABC+∠ACB=180°
∴∠CBE+∠BCF=∠A+∠ACB +∠A+∠ABC=∠A+180°
又∵BD CD分别是∠CBE、∠BCF的平分线
∴∠CBD+∠BCD=1/2*（∠CBE+∠BCF）=1/2*（∠A+180°）=∠A/2+90°
∴∠D=180°-∠A/2-90°=90°-∠A/2

Answer 2

∠BDC=180-1/2(∠CBE+∠BCF)
=180°-1/2(2∠A+∠ABC+∠ACB)
=180°-1/2*∠A-1/2(∠A+∠ABC+∠ACB)
=180°-1/2*∠A-1/2*180°
=180°-1/2∠A-90°
=90°-∠A/2

Answer 3

∠BDC=90°-1/2∠A