Solution :
Given : A trapezium ABC whose diagonals AC and BD intersect each other at O and AB || DC.
To Prove : \(OA\over OC\) = \(OB\over OD\)
Proof : In triangle DOC and AOB,
AB || DC and AC is transversal, then
\(\angle\) DCO = \(\angle\) OAB (Alternate angles)
\(\angle\) ODC = \(\angle\) OBA (Alternate angles)
\(\angle\) DOC = \(\angle\) AOB (vertically opposite angles)
Hence, By AAA similarity,
\(\triangle\) DOC ~ \(\triangle\) BOA
\(\implies\) \(OA\over OC\) = \(OB\over OD\)