Question

In a quadrilateral $A BC D$ , diagonals $A C$ and $B D$ intersect each other at $O$ such that $\frac{A O}{BO} = \frac{CO}{D O}$ as shown in the given figure. Prove that $A BC D$ is a trapezium.

Approach

Using Basic Proportionality Theorem (BPT) converse: in $△ A B D$ , since $\frac{A O}{BO} = \frac{CO}{D O}$ , line through $O$ parallel to $A B$ implies $D C ∥ A B$ , hence trapezium.

Step-by-step working

1
Given $\frac{A O}{BO} = \frac{CO}{D O}$
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2
In $△ A B D$ and $△ B D C$ , using BPT converse
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3
$A B ∥ D C$
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Hence $A BC D$ is a trapezium
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5
Diagram
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Concepts used

similar trianglestrapeziumbasic proportionality

Approach

Step-by-step working

Concepts used

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