Diagonal characterization of parallelograms

Step 1 (parallelogram implies bisecting diagonals). Assume $ABCD$ is a parallelogram. Because opposite sides are parallel and equal, we have $AB = CD$, and the alternate interior angles cut by the two diagonals satisfy $\angle OAB = \angle OCD$ and $\angle OBA = \angle ODC$. By the Angle-Side-Angle criterion, $\triangle AOB \cong \triangle COD$, so corresponding sides give $OA = OC$ and $OB = OD$.

Step 2 (bisecting diagonals imply congruent opposite triangles). Conversely, assume $OA = OC$ and $OB = OD$. Because vertical angles at the intersection are equal ($\angle AOB = \angle COD$), the Side-Angle-Side criterion yields $\triangle AOB \cong \triangle COD$, which gives $\angle OAB = \angle OCD$. By the exact same Side-Angle-Side argument on the other pair of vertical angles ($\angle AOD = \angle COB$), we obtain $\triangle AOD \cong \triangle COB$, which gives $\angle OAD = \angle OCB$.

Step 3 (equal alternate interior angles force parallel sides). The equalities $\angle OAB = \angle OCD$ and $\angle OAD = \angle OCB$ state that the alternate interior angles formed by the diagonal transversal with both pairs of opposite sides are equal, forcing $AB \parallel CD$ and $AD \parallel BC$. Thus $ABCD$ is a parallelogram.