Sections AC and BD intersect in the middle O of segment AC, angle BCO = angle DAO. Prove that triangle BOA = triangle DOC

Given: О – middle AC angle BCO = angle DAO
Prove: triangle BOA = triangle DOC Proof: angle BCO = angle DAO (by condition) AO = OC (by condition) angle AOD = angle BOS, because vertical> from these three proofs it follows that triangle DOA = triangle BOC along the side and 2m adjacent angles> it follows from this DO = BO, AO = OC (by conv), angle AOB = angle DOC, mk vertical> it follows that triangle BOA = triangle DOC on 2 sides and the angle between them. Q.E.D.