C-07-Q062mediumsingle_mcqFrom ∠ABO=∠CDO\angle ABO = \angle CDO∠ABO=∠CDO being alternate angles, it follows that—aAB⊥CDAB \perp CDAB⊥CDbABABAB and CDCDCD are parallel and equalcAC=BDAC = BDAC=BDd△ABC\triangle ABC△ABC is right-angled