题目如图,在半径为555的⊙\odot⊙中,ABABAB为⊙O\odot O⊙O的直径,OD⊥OD\botOD⊥弦ACACAC交⊙O\odot O⊙O于DDD,垂足是HHH,BDBDBD交ACACAC于EEE,过点EEE作EF⊥EBEF\bot EBEF⊥EB交⊙O\odot O⊙O于FFF,且EF=EBEF=EBEF=EB,连接OFOFOF,AFAFAF,BFBFBF.(1)(1)(1)求证:∠OFE=∠ODE\angle OFE=\angle ODE∠OFE=∠ODE;(2)(2)(2)若EH=1EH=1EH=1,求AFAFAF的长.知识点:垂径定理章节:图形的性质 / 圆