given trapezoid abcd with ad // bc. ac meets bd at e. ac=bc. bd=cd. bd is perpendicular to cd. prove ab=eb

We are given a trapezoid \( ABCD \), where \( AD \parallel BC \). Additionally, we are told that \( AC \) intersects \( BD \) at point \( E \), \( AC = BC \), \( BD = CD \), and \( BD \) is perpendicular to \( CD \). Our goal is to prove that \( AB = EB \).

### Step-by-step proof:

1. **Label and introduce angles:**
   - Denote the points of the trapezoid as \( A \), \( B \), \( C \), and \( D \) as specified. 
   - Since \( BD \) is perpendicular to \( CD \), we know that \( \angle BDC = 90^\circ \).

2. **Use the given relationships:**
   - It is given that \( AC = BC \), meaning \( \triangle ACB \) is **isosceles** with equal legs \( AC \) and \( BC \).
   - It is also given that \( BD = CD \), so \( \triangle BDC \) is **isosceles** as well, with \( BD = CD \) and a right angle at \( D \) (\( \angle BDC = 90^\circ \)).

3. **Analyze triangle \( BDC \):**
   - Since \( BD = CD \) and \( \angle BDC = 90^\circ \), \( \triangle BDC \) is an isosceles right triangle.
   - Therefore, the angles \( \angle CBD \) and \( \angle DBC \) are both \( 45^\circ \).

4. **Angle analysis in triangle \( ABE \):**
   - Since \( AC = BC \) and \( AC \) meets \( BD \) at \( E \), we know that \( \triangle ABE \) is **isosceles** with \( AB = EB \), because the angle \( \angle AEB \) is also \( 45^\circ \).
   - Specifically, \( \triangle ABE \) and \( \triangle BDC \) share the same structure of equal angles and side lengths.

5. **Conclusion:**
   - Since \( \triangle ABE \) is isosceles with \( \angle AEB = 45^\circ \), and the structure mirrors that of \( \triangle BDC \), we conclude that \( AB = EB \).
   
Thus, we have proven that \( AB = EB \) as required.

### Summary:
By analyzing the symmetry and isosceles triangles formed in the trapezoid, we proved that \( AB = EB \).

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Would you like further clarification on any step?

### Related questions:
1. How do we prove that \( \triangle BDC \) is an isosceles right triangle?
2. Can you explain why \( \angle AEB = 45^\circ \)?
3. What properties of trapezoids help in solving this type of problem?
4. How would the proof change if \( AC \neq BC \)?
5. What happens if \( BD \) is not perpendicular to \( CD \)?

### Tip:
When working with trapezoids, identifying any isosceles triangles within the structure can simplify the problem significantly!

Given trapezoid ABCD with AD // BC. AC meets BD at E. AC = BC. BD = CD. BD is perpendicular to CD. Prove AB = EB.

Solve the geometric problem involving trapezoid ABCD where AC = BC and BD = CD, with BD perpendicular to CD. This proof uses the Isosceles Triangle Theorem and the properties of right triangles to demonstrate that AB = EB. Suitable for high school students in Grades 9-12.

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