C-08-Q101mediumsingle_mcqAdding the two relations gives ∠BOC=⋯\angle BOC = \dotsb∠BOC=⋯a∠BAC\angle BAC∠BACb2∠BAC2\angle BAC2∠BACc12∠BAC\frac{1}{2}\angle BAC21∠BACd∠BAO+∠CAD\angle BAO + \angle CAD∠BAO+∠CAD