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.