The diagonals of a quadrilateral ABCD intersect each other at the point O such that \(AO\over BO\) = \(CO\over DO\). Show that ABCD is a trapezium.

Solution :

Given : A quadrilateral ABCD, in which the diagonals AC and BD intersect each other at O such that trapezium \(AO\over BO\) = \(CO\over DO\).

To  Prove : ABCD is a trapezium.

Construction : Draw EO || BA, meeting AD in E.

Proof : In triangle ABD, EO || BA

By basic proportionality theorem,

\(DE\over EA\) = \(DO\over OB\)                    …………(1)

But Given that, \(AO\over BO\) = \(CO\over DO\) \(\implies\)  \(DO\over BO\) = \(CO\over AO\)               …………..(2)

From (1) and (2), we get

\(DE\over EA\) = \(CO\over OA\)

By converse of basic proportionality theorem,

\(\implies\) EO || DC

But by construction, EO || BA

\(\therefore\)  DC || BA

Hence, ABCD is a trapezium.

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