题目如图,ABABAB是⊙O\odot O⊙O的直径,弦CDCDCD交ABABAB于点PPP,AP=2AP=2AP=2,BP=6BP=6BP=6,∠APC=30∘\angle APC=30^{\circ}∠APC=30∘,则CDCDCD的长为( )A.15\sqrt{15}15B.252\sqrt{5}25C.2152\sqrt{15}215D.888知识点:垂径定理章节:图形的性质 / 圆