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The problem describes a rectangle \(ABCD\) with the following dimensions and relationships:

- Sides: \(AB = CD = 34 \, \text{cm}\)
- \(E\) is a point on side \(CD\) such that \(CE = 9 \, \text{cm}\) and \(ED = 25 \, \text{cm}\) (since \(CD = 34 \, \text{cm}\), \(9 + 25 = 34 \, \text{cm}\)).
- \(AE\) is perpendicular to diagonal \(EB\).
  
The goal is to find the length of \(BC\).

### Solution approach:

1. **Understand the rectangle**: In any rectangle, opposite sides are equal, and all angles are 90 degrees.

2. **Geometrical setup**: 
   - \(E\) is a point on \(CD\), and a perpendicular line \(AE\) from \(A\) meets \(EB\).
   - The problem suggests a geometric configuration where triangles or Pythagoras' theorem may come into play, given that \(AE \perp EB\).

3. **Use of Pythagoras' Theorem**:
   - In triangle \(AEB\), since \(AE\) is perpendicular to \(EB\), it forms a right triangle. We can apply the Pythagorean theorem to solve for lengths in this triangle.

4. **Breaking down diagonals**: 
   - Use the diagonal properties of the rectangle to help relate the sides.

Let's solve step-by-step:

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A rectangle ABCD has sides AB = CD = 34 cm. E is a point on CD such that CE = 9 cm and ED = 25 cm. AE is perpendicular to EB. What is the length of BC?

Solve for the length of BC in a rectangle ABCD with AB = CD = 34 cm and E on CD, where AE is perpendicular to EB. This problem uses geometric properties and the Pythagorean theorem. Suitable for Grades 9-12.

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