题目如图,△ABC\triangle ABC△ABC内接于⊙O\odot O⊙O,ABABAB是⊙O\odot O⊙O的直径,点EEE在⊙O\odot O⊙O上,点CCC是BE^\widehat {BE}BE的中点,AE⊥CDAE\bot CDAE⊥CD,垂足为点DDD,DCDCDC的延长线交ABABAB的延长线于点FFF.(1)(1)(1)求证:CDCDCD是⊙O\odot O⊙O的切线;(2)(2)(2)若CD=3CD=\sqrt{3}CD=3,∠ABC=60∘\angle ABC=60^{\circ}∠ABC=60∘,求线段AFAFAF的长.知识点:切线的判定章节:图形的性质 / 圆