CBSE Class 10 Maths (Basic) 2025 · Paper 430/2/1 · Question 33

**SAS criterion:** if one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, the triangles are similar. **Proof:** from $OA\cdot OC=OB\cdot OD$, $\frac{OA}{OB}=\frac{OD}{OC}$, and $\angle AOD=\angle BOC$ (vertically opposite angles), so $△OAD\sim △OBC$ by SAS. Hence $\angle OAD=\angle OBC$, which are alternate angles on the transversal $AB$, so $AD\parallel CB$. Hence proved.