Question

Given: DF=EG

An Image shows a line with dots on them the first dot is D the second dot is E the third dot is F and the fourth dot is G

Prove: DE=FG

Statements Reasons
DF=EG ________
DF=DE+EF Segment Addition Postulate
EG=EF+FG Segment Addition Postulate
DE+EF=EF+FG _____________
DE=FG _________


A. Subtraction Property of Equality
B. Given
C. Substitution Property of Equality
D. Segment Congruence Postulate
E. Definition of midpoint

Answer 1

To prove that \( DE = FG \) given \( DF = EG \) and using the Segment Addition Postulate, we can fill in the statements and reasons as follows:

**Proof:**

| Statements | Reasons |
|-------------------------------|-------------------------------------|
| 1. \( DF = EG \) | B. Given |
| 2. \( DF = DE + EF \) | Segment Addition Postulate |
| 3. \( EG = EF + FG \) | Segment Addition Postulate |
| 4. \( DE + EF = EF + FG \) | C. Substitution Property of Equality|
| 5. \( DE = FG \) | A. Subtraction Property of Equality |

Here’s how the proof works:

1. From the given information, we start with \( DF = EG \).
2. According to the Segment Addition Postulate, we can write \( DF \) as \( DE + EF \).
3. Similarly, we can express \( EG \) as \( EF + FG \) using the Segment Addition Postulate.
4. Since both \( DF \) and \( EG \) are equal, we can substitute \( DE + EF \) for \( DF \) and \( EF + FG \) for \( EG \). This gives us \( DE + EF = EF + FG \).
5. Finally, we can apply the Subtraction Property of Equality, which allows us to subtract \( EF \) from both sides, resulting in \( DE = FG \).

Thus, we have proven that \( DE = FG \).