To show that \( AD = BE \), given the conditions \( AB = DE \) and \( BC = CD \), let's proceed step by step.

### Step 1: Assign Variables to the Lengths
Let:
- \( AB = x \),
- \( BC = y \),
- \( CD = y \) (since \( BC = CD \)),
- \( DE = x \) (since \( AB = DE \)).

### Step 2: Express \( AD \) and \( BE \)
Now, let's express the lengths of \( AD \) and \( BE \):
- \( AD \) is the total length from \( A \) to \( D \), which consists of two parts: \( AB + BC \). Thus:
  \[
  AD = AB + BC = x + y
  \]
- Similarly, \( BE \) is the total length from \( B \) to \( E \), which consists of two parts: \( BC + CD \) (since \( BC = CD \), these two segments are equal). So:
  \[
  BE = BC + DE = y + x
  \]

### Step 3: Conclusion
Since \( AD = x + y \) and \( BE = y + x \), and addition is commutative (i.e., \( x + y = y + x \)), we can conclude that:
\[
AD = BE
\]

Thus, it is proven that \( AD = BE \).

Would you like more details or have any further questions?

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Here are 5 related questions to expand on this concept:
1. What happens if \( AB \neq DE \) but \( BC = CD \)?
2. How would you approach proving this geometrically?
3. Can you generalize this result to different shapes or configurations?
4. What role does symmetry play in this type of proof?
5. How would this proof change if additional constraints were given?

**Tip:** When solving geometric problems involving line segments, always break them down into smaller parts and use algebra to simplify the relations.

Learn how to prove that AD = BE given the conditions AB = DE and BC = CD. This solution uses geometric principles and algebra to express line segment relationships, applying the commutative property of addition. Ideal for students in Grades 7-9.

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