在△ABC中,若sin(B/2)=sin(A/2)sin(C/2),求证:a+c=3b

sinB
= 2sin(B/2)cos(B/2)
= 2sin(A/2)sin(C/2)cos(B/2)
= sin(A/2)[sin((B+C)/2) - sin((B-C)/2)]
= sin(A/2)[cos(A/2)- sin((B-C)/2)]
= (1/2)sinA - cos[(B+C)/2]sin[(B-C)/2]
= (1/2)(sinA-sinB +sinC)
3sinB = sinA + sinC
根据正弦定理 a=2RsinA,b=2RsinB,c=2RsinC
a+c=3