https://file.aiquickdraw.com/mathbot/file/1726869467236yzo1w15y.jpg

The image shows a proof table with the given information:

**Given**:  
- \( AC \cong DF \)  
- \( BC \cong DE \)

**To Prove**:  
- \( AB \cong EF \)

### Statements and Reasons (so far):

1. \( AC \cong DF \), \( BC \cong DE \) – **Reason**: Given  
2. \( AB + BC = AC \) – **Reason**: Segment Addition Postulate  
3. \( AB + DE = DF \) – **Reason**: Substitution (since \( BC \cong DE \) and \( AC \cong DF \))

To complete the proof, let's continue step by step:

4. \( AB + DE = AB + BC \) – **Reason**: \( BC \cong DE \)  
5. \( AB = EF \) – **Reason**: By substitution and rearranging, since \( AB + DE = DF \), and \( EF = AB \).

The reasoning involves breaking down the segment addition postulate and substituting congruent segments based on the given information.

### Do you have any questions or would you like further details on this?

### Follow-up questions:
1. Why is the Segment Addition Postulate used in this proof?
2. How does substitution of congruent segments work in this proof?
3. Could this proof also be solved using a different geometric property?
4. What is the significance of proving \( AB \cong EF \) in the context of parallel lines?
5. Can you extend this proof for a scenario where the lines are not parallel?

### Tip:
When working with geometric proofs, always verify that each step logically follows from the previous ones, either through definitions, postulates, or theorems.

Given AC ≅ DF, BC ≅ DE, prove AB ≅ EF using a geometric proof.

Solve the geometric proof with the given conditions AC ≅ DF and BC ≅ DE. Learn how to prove AB ≅ EF using the Segment Addition Postulate and Substitution Property, step-by-step.

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