已知△ABC中,2B=A+C,b²=ac,证明：△ABC为等边三角形,

Question

已知△ABC中,2B=A+C,b²=ac,证明：△ABC为等边三角形,

Answer 1

证明:
∵2B=A+C,且A+B+C=180
∴B=60
由正弦定理a/sinA=b/sinB=c/sinC=k
得a=ksinA,b=ksinB,c=ksinC 代入b²=ac
(ksinB)²=ksinA*ksinC
(sinB)²=sinA*sinC
sinA*sinC=3/4
积化和差
sinAsinC=-[cos(A+C)-cos(A-C)]/2
∵A+C=2B=120
∴sinAsinC=-[cos120-cos(A-C)]/2=3/4
∴cos(A-C)=1
∴A=C.又B=60
∴△ABC为等边三角形