In a quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at $O$ such that $\dfrac{AO}{BO} = \dfrac{CO}{DO}$. Prove that $ABCD$ is a trapezium.
Step-by-Step Solution
Key Concept: Rewrite $\dfrac{AO}{OC} = \dfrac{BO}{OD}$. Draw $OE \parallel AB$ meeting $AD$ at $E$. Show $AB \parallel CD$.
Rearrange given ratio: $\dfrac{AO}{OC} = \dfrac{BO}{OD}$. [0.5 Mark]
Draw line $OE \parallel AB$ meeting $AD$ at $E$. In $\Delta DAB$, $OE \parallel AB \Rightarrow \dfrac{DE}{EA} = \dfrac{DO}{OB}$ (by BPT). (1) [1.0 Mark]
Given $\dfrac{DO}{OB} = \dfrac{CO}{OA} \Rightarrow \dfrac{DE}{EA} = \dfrac{CO}{OA}$. By converse of BPT in $\Delta ADC$, $OE \parallel CD$. [1.0 Mark]
Since $OE \parallel AB$ and $OE \parallel CD$, $AB \parallel CD$. Hence $ABCD$ is a trapezium. Proved! [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Construction $OE \parallel AB$: 0.5 Mark
Applying BPT to get $DE/EA = DO/OB$: 1.0 Mark
Applying converse of BPT to show $OE \parallel CD$: 1.0 Mark
Concluding $AB \parallel CD \Rightarrow ABCD$ is a trapezium: 0.5 Mark
Correct Answer: