ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that \(AO\over BO\) = \(CO\over DO\).

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

\(DO\over OB\) = \(CO\over OA\)

\(\implies\)    \(AO\over BO\) = \(CO\over DO\)