Understanding the Trapezium and Its Properties
In trapezium ABCD, where AB is parallel to DC, we need to prove that the ratio of the segments created by the intersecting diagonals is equal, specifically AO/BO = CO/DO.

Diagonals in a Trapezium
- When the diagonals AC and BD intersect at point O, they create four segments: AO, BO, CO, and DO.
- Since AB || DC, the trapezium has properties related to similar triangles.

Applying the Basic Proportionality Theorem
- The Basic Proportionality Theorem states that if a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.
- In trapezium ABCD, if we drop perpendiculars from points A and B to line DC, we can form two triangles: triangle AOD and triangle BOC.

Establishing Proportions
- Since AB || DC, triangles AOD and BOC are similar by the AA criterion (Angle-Angle).
- This similarity gives us the proportion:
AO/BO = OD/OC
- Similarly, we can consider triangles COD and AOB, which will also yield:
CO/DO = AO/BO

Conclusion
- From the equal ratios derived from the similarity of triangles, we conclude that:
AO/BO = CO/DO
- This proves that the ratios of the segments formed by the diagonals in trapezium ABCD maintain equality as required.
Thus, you have shown that in trapezium ABCD with diagonals intersecting at O, the segments are proportionally equal!