The segments AC and BC are halved by the intersection. Prove that triangle ABC = triangle CDA.

Given:
segments AC, BD;
segments AC and BD intersect at point O;
AO = OС;
BO = OD.
Prove that triangle ABC = triangle CDA.
Evidence:
1) Triangle AOD = triangle BOC on two sides and the angle between them, since AO = OC, BO = OD and angle AOD = angle BOC (vertical). Then AD = BC.
2) Triangle AOB = COD on two sides and the angle between them, since AO = OC, BO = OD and angle AOB = angle DOC (vertical). Then AB = CD.
3) Triangle ABC = triangle CDA on three sides, since AD = BC, AB = CD and AC are common.
Proven.