题目如图,ABABAB是⊙O\odot O⊙O的直径,点FFF,CCC是⊙O\odot O⊙O上两点,且AF^=FC^=CB^\widehat {AF}=\widehat {FC}=\widehat {CB}AF=FC=CB,连接ACACAC,AFAFAF,过点CCC作CD⊥AFCD\bot AFCD⊥AF交AFAFAF延长线于点DDD,垂足为DDD.(1)(1)(1)求∠BAC\angle BAC∠BAC的度数;(2)(2)(2)求证:CDCDCD是⊙O\odot O⊙O的切线;(3)(3)(3)若CD=23CD=2\sqrt{3}CD=23,求⊙O\odot O⊙O的半径.知识点:切线的判定章节:图形的性质 / 圆