The segments BD and CA meet at point O. It is known that BC is parallel to AD and BC = DA.

The segments BD and CA meet at point O. It is known that BC is parallel to AD and BC = DA. Prove that triangle ABO = triangle DOC.

Since, according to the condition, AD is parallel to BC and AD = BC, then the angle CBO = ADO, the angle BCO = DAO as criss-crossing angles at the intersection of parallel straight lines BC and AD secant AC and BD.

Then the triangles BOC and AOD are equal in side and two adjacent angles, which means AO = CO, BO = DO.

In triangles ABO and DOC, the angle BOA = COD as vertical angles, then the triangles ABO and DOC are equal on two sides and the angle between them, which was required to be proved.