题目已知,ABABAB为⊙O\odot O⊙O的直径,弦CDCDCD交ABABAB于点EEE,连接OCOCOC,ADADAD,∠BOC=2∠BAD\angle BOC=2\angle BAD∠BOC=2∠BAD.(1)(1)(1)如图111,求证:AB⊥CDAB\bot CDAB⊥CD;(2)(2)(2)如图222,过点BBB作BFBFBF∥OCOCOC,交⊙O\odot O⊙O于点FFF,点GGG为BFBFBF的中点,连接OGOGOG,求证:CE=OGCE=OGCE=OG;(3)(3)(3)如图333,在(2)的条件下,延长COCOCO交ADADAD于点HHH,若BE=BFBE=BFBE=BF,△AOH\triangle AOH△AOH的面积为525\sqrt{2}52,求线段DHDHDH的长.知识点:垂径定理章节:图形的性质 / 圆