题目如图,OAOAOA、OBOBOB、OCOCOC均为⊙O\odot O⊙O半径,∠AOC=12∠AOB∠AOC=\frac{1}{2}∠AOB∠AOC=21∠AOB,下列结论不正确的是( )A.AB=2BCAB=2BCAB=2BCB.∠ACB=2∠ABC\angle ACB=2\angle ABC∠ACB=2∠ABCC.AB^=2AC^\widehat {AB}=2\widehat {AC}AB=2ACD.∠ACB=∠AOC\angle ACB=\angle AOC∠ACB=∠AOC知识点:角的大小比较章节:未标注