题目如图,已知ABABAB、ACACAC是半径为111的⊙O\odot O⊙O的两条弦,且AB=ACAB=ACAB=AC,BOBOBO的延长线交ACACAC于点DDD,连接OAOAOA、OCOCOC.(1)(1)(1)证明:∠ABO=∠ACO\angle ABO=\angle ACO∠ABO=∠ACO；(2)(2)(2)试求(ADOD)2−OAOD(\frac{AD}{OD})^{2}-\frac{OA}{OD}(ODAD)2−ODOA的值；(3)(3)(3)记△AOB\triangle AOB△AOB、△AOD\triangle AOD△AOD、△COD\triangle COD△COD的面积分别为S1S_{1}S1、S2S_{2}S2、S3S_{3}S3,若S22=2S1•S3S_{2}^{2}=2S_{1}•S_{3}S22=2S1•S3,求ODODOD的长.知识点：圆的综合题章节：图形的性质 / 圆
