Solution :
Given : A trapezium ABCD, in which the diagonals AC and BD intersect each other at O.
To Prove : \(AO\over BO\) = \(CO\over DO\)
Construction : Draw EF || BA || CD, meeting AD in E.
Proof : In triangle ABD, EF || AB
By basic proportionality theorem,
\(DO\over OB\) = \(DE\over AE\) …………(1)
In triangle CDA, EO || DC,
By basic proportionality theorem,
\(CO\over OA\) = \(DE\over AE\) …………..(2)
From (1) and (2), we get
\(DO\over OB\) = \(CO\over OA\)
\(\implies\) \(AO\over BO\) = \(CO\over DO\)