题目已知⊙O\odot O⊙O为△ABC\triangle ABC△ABC的外接圆,AB=BCAB=BCAB=BC.过AAA作COCOCO的垂线交COCOCO延长线于点DDD,则下列选项一定成立的是( )A.∠BCA=∠DCA\angle BCA=\angle DCA∠BCA=∠DCAB.∠DAC=2∠BAC\angle DAC=2\angle BAC∠DAC=2∠BACC.AB>2ADAB \gt 2ADAB>2ADD.4AB2<AD2+CD24AB^{2} \lt AD^{2}+CD^{2}4AB2<AD2+CD2知识点:切线的判定章节:图形的性质 / 圆