Question

State the SAS similarity criterion: if one angle of a triangle equals one angle of another triangle and the sides including these angles are proportional, the triangles are similar. Given $OA\cdot OC=OB\cdot OD$, rewrite as $\frac{OA}{OB}=\frac{OD}{OC}$. Also, $\angle AOD=\angle BOC$ (vertically opposite). By SAS similarity, $△OAD\sim △OBC$, so $\angle A=\angle B$. Since these are alternate interior angles for lines $AD$ and $CB$ with transversal $AB$, we have $AD\parallel CB$.