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°

∠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

∠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°

在三角形AB