Question

已知△ABC的两条外角平分线BD与CD交于点D。求证∠D+1/2∠A=90°

Answer 1

∠CBD=1/2(∠A+∠C)

∠DCB=1/2(∠A+∠B)

∠D=180-∠CBD-∠DCB=180-1/2(∠A+∠C)- 1/2(∠A+∠B)=1/2(∠B+∠C)

因∠A+∠B+∠C=180

故∠D+1/2∠A=1/2(∠B+∠C)+ 1/2∠A=1/2(∠A+∠B+∠C)=1/2*180=90

Answer 2

∠D=180°-∠DBC-∠BCD
∠DBC=1/2(∠ABC的外角)=1/2(∠A+∠ACB)
∠BCD=1/2(∠ACB的外角)=1/2(∠A+∠ABC)
故∠D=180°-∠DBC-∠BCD
=180°-1/2(∠A+∠ACB) -1/2(∠A+∠ABC)
=180°-∠A-1/2(∠ACB+∠ABC)
=180°-∠A-1/2(180°-∠A)
=90°-1/2*∠A
故∠D+1/2∠A=90°

Answer 3

在三角形ABC中
∠A+∠ABC+∠ACB=180

在三角形BCD中
∠D+（180-∠ABC）/2+（180-∠ACB）/2=180

联立得∠D+1/2∠A=90

Answer 4

证明：在△ABC中
∠ABC+∠ACB=180°－∠A
∠BDC=180°-(∠CBD+∠BCD)
∵BD、CD为∠ABC，∠ACB的外角∠EBC和∠FCB的平分线
∴∠EBD=∠CBD=1/2∠CBE
∠BCD=FCD＝1/2∠BCF
∴∠BDC=180°-(∠CBD+∠BCD)
=180°-1/2(∠CBE＋∠BCF)
∵∠CBE=180°-∠ABC，∠BCF=180°-∠ACB
∴∠BDC=180°-1/2(180°-∠ABC+180°-∠ACB)
=1/2(∠ABC+∠ACB)
＝1/2(180°-∠A)
∴∠D+1/2∠A＝90°

Answer 5

那我就精练点呵.∠D=180°-∠DBC-∠BCD
∠DBC+∠BCD={360-[180-∠A]}/2
∠D=90=A/2
所以∠D+∠A/2=90

Answer 6

这是个错误的问题