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!