题目如图,PAPAPA,PBPBPB切⊙O⊙O⊙O于AAA,BBB两点,CDCDCD切⊙O⊙O⊙O于点EEE交PAPAPA,PBPBPB于CCC,DDD,若⊙O⊙O⊙O的半径为rrr,△PCD\triangle PCD△PCD的周长为3r3r3r,连接OAOAOA,OPOPOP,则OAPA\dfrac{OA}{PA}PAOA的值是()(\quad)()A.21313\dfrac{2}{13}\sqrt{13}13213B.125\dfrac{12}{5}512C.32\dfrac{3}{2}23D.23\dfrac{2}{3}32知识点:切线长定理章节:图形的性质 / 圆