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