摘要：解:由已知2lgsinB=lgsinA+lgsinC. 得lg(sinB)2=lg(sinA·sinC). ∴sin2B=sinA·sinC. 设l1:a1x+b1y+c1=0.l2:a2x+b2y+c2=0． ∵===. =. ===. ∴==.l1与l2重合．答案:D

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