Diagonals AC and BD of trapezium ABCD with AB||DC intersect each other at point O. Show that $\frac{OA}{OC} = \frac{OB}{OD}$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:27 · grounding rag
Model Answer
In △AOB and △COD:
∠OAB = ∠OCA (Alternate interior angles, since AB ∥ DC)
∠OBA = ∠ODC (Alternate interior angles, since AB ∥ DC)
∴ △AOB ~ △COD (AA similarity criterion)
Therefore, $\dfrac{OA}{OC} = \dfrac{OB}{OD}$ (Corresponding sides of similar triangles)
Hence proved.
Source: Triangles, Exercise 6.3 Q.3, Chapter 6
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Explanation
- The key step is identifying the two triangles to compare: △AOB and △COD (not △AOB and △DOC — order of vertices matters for correct correspondence).
- AB ∥ DC gives alternate interior angles for both pairs: ∠OAB = ∠OCA and ∠OBA = ∠ODC. Two pairs of equal angles → AA criterion.
- Once similarity is established, write the ratio from corresponding vertices in order: A↔C, O↔O, B↔D, giving OA/OC = OB/OD directly.
- Examiners expect you to name the similarity criterion explicitly (AA) and state the reason for the angle equality.