7.2 SAS Axiom & ASA Theorem

Given: $AB \parallel CD$, $OA = OD$.

Consider $\Delta AOB$ and $\Delta DOC$.

1. $\angle BAO = \angle CDO$ (Alternate Interior Angles, since $AB \parallel CD$).

2. $OA = OD$ (Given).

3. $\angle AOB = \angle DOC$ (Vertically Opposite Angles).

By ASA Criterion, $\Delta AOB \cong \Delta DOC$.