The problem you're working on involves proving a property related to diagonals of a trapezium. Let's break down the steps clearly.

Problem Recap:

• You are given a trapezium $ABCD$, where $AB \parallel DC$.
• The diagonals $AC$ and $BD$ intersect at a point $O$.
• You are asked to prove: $\frac{AO}{BO} = \frac{CO}{DO}$

This is a standard problem in geometry related to trapeziums (trapezoids) and can be solved using similar triangles.

Step-by-Step Explanation:

1. Label the Diagram:

   • Trapezium $ABCD$ has parallel sides $AB \parallel DC$.
   • The diagonals $AC$ and $BD$ intersect at point $O$.
   • We need to prove that the ratio of segments created by the diagonals is equal, i.e., $\frac{AO}{BO} = \frac{CO}{DO}$.

2. Identify Similar Triangles: Since $AB \parallel DC$, the angles formed by the diagonals and the parallel sides will be equal by alternate interior angles.

   • Consider $\triangle AOB$ and $\triangle COD$. These two triangles share the following angles:
     • $\angle OAB = \angle OCD$ (because $AB \parallel DC$, alternate interior angles).
     • $\angle OBA = \angle ODC$ (again, due to $AB \parallel DC$, alternate interior angles).
     • $\angle AOB = \angle COD$ (vertically opposite angles at the intersection point $O$).

   Therefore, by AA similarity (two corresponding angles are equal), $\triangle AOB \sim \triangle COD$.

3. Use of Proportionality in Similar Triangles: Since $\triangle AOB \sim \triangle COD$, corresponding sides of these triangles will be proportional. This gives us the relationship: $\frac{AO}{BO} = \frac{CO}{DO}$ This directly proves the required result.

Conclusion:

By proving that the triangles $AOB$ and $COD$ are similar, we establish the proportionality of the corresponding sides, which leads to the desired ratio equality: $\frac{AO}{BO} = \frac{CO}{DO}$

Would you like further clarification on any part? Here are some additional questions to deepen understanding:

1. What properties of parallel lines and alternate interior angles did we use in this proof?
2. How does the concept of similar triangles apply to problems involving diagonals of trapeziums?
3. Can you identify other geometric properties that hold when diagonals intersect in quadrilaterals?
4. How would the problem change if $AB$ was not parallel to $DC$?
5. Can this proportionality rule be applied to other quadrilaterals with parallel sides?

Tip:

When working with trapeziums, always check for parallel sides first, as they often lead to useful angle relationships and the possibility of similar triangles.