The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\frac{AO}{OC} = \frac{BO}{OD}$. Show that quadrilateral ABCD is a trapezium.
Generated by claude-sonnet-4-6 · 2026-06-15 10:26 · grounding rag
Model Answer
Given: Diagonals of quadrilateral ABCD intersect at O such that $\dfrac{AO}{OC} = \dfrac{BO}{OD}$.
To prove: ABCD is a trapezium (i.e., AB ∥ DC).
Through O, draw EO ∥ AB, meeting AD at E.
In △ADB, EO ∥ AB (by construction), so by Basic Proportionality Theorem:
$$\frac{AE}{ED} = \frac{BO}{OD} \quad \cdots (1)$$
Given: $\dfrac{AO}{OC} = \dfrac{BO}{OD} \quad \cdots (2)$
From (1) and (2): $\dfrac{AO}{OC} = \dfrac{AE}{ED}$
In △ACD, E is on AD and O is on AC such that $\dfrac{AE}{ED} = \dfrac{AO}{OC}$.
By the converse of Basic Proportionality Theorem, EO ∥ DC.
But EO ∥ AB (by construction). Therefore, AB ∥ DC.
Hence, ABCD is a trapezium. [Proved]
Source: Chapter 6, Section 6.3 (Theorem 6.1 and Theorem 6.2)
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Explanation
- The key tool here is Theorem 6.1 (BPT) and its converse (Theorem 6.2).
- The trick is to draw an auxiliary line EO ∥ AB through O to create a triangle where BPT applies, then use the given ratio to show EO ∥ DC as well, forcing AB ∥ DC.
- Examiners expect you to clearly state which theorem you are applying at each step. Don't skip the construction — it's essential.
- "ABCD is a trapezium" means one pair of opposite sides is parallel, so proving AB ∥ DC is sufficient.