✔ 最佳答案
三角形ABC中,a=(b+c)Sin(A/2),則此三角形為?
Sol
a/(b+c)=SinA/(SinB+SinC)=2Sin(A/2)Cos(A/2)/(SinB+SinC)
2Sin(A/2)Cos(A/2)/(SinB+SinC)=Sin(A/2)
Sin(A/2)<>0
2Cos(A/2)/(SinB+SinC)=1
2Cos(A/2)=SinB+SinC
Set P=B+C,Q=B-C
P+Q=2B,P-Q=2C
A=π-P
2Cos(A/2)=SinB+SinC
2Cos(π/2-P/2)=Sin(P/2+Q/2)+Sin(P/2-Q/2)
2Sin(P/2)=2Sin(P/2)Cos(Q/2)
Sin(P/2)[1-Cos(Q/2)]=0
(1) Sin(P/2)=0
Sin(π/2-A/2)=0
Cos(A/2)=0 (不合)
(2) Cos(Q/2)=1
Q/2=0
B-C =0
B=C
等腰三角形