Question

三角形ABC中,若asinA+bsinB=csinC+asinB,且c=2bcosA

Answer 1

由c=2bcosA可得,c/b=sinC/sinB=2cosA,sinC=sin(A+B)=sinAcosB+sinBcosA,即sinAcosB-sinBcosA=0,sin(A-B)=0,A=B,a=b,则asinA+bsinB=csinC+asinB可化简为:asinA=csinC,a/c=sinA/sinC=sinC/sinA,

(sinA)^2=(sinC)^2,sinA=sinC,A=C,故三角形ABC为等边三角形.