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Sagot :
Here's a detailed two-column proof for the given problem.
| Statements | Reasons |
|-----------------------------------------------|----------------------------------------------|
| 1. [tex]\( AB = CD \)[/tex] | Given |
| 2. [tex]\( BC = DE \)[/tex] | Given |
| 3. [tex]\( AB + BC = CD + BC \)[/tex] | Addition property of equality |
| 4. [tex]\( CD + BC = CD + DE \)[/tex] | Substitution property of equality |
| 5. [tex]\( AB + BC = AC \)[/tex] | Segment addition (AB + BC = AC) |
| 6. [tex]\( CD + DE = CE \)[/tex] | Segment addition (CD + DE = CE) |
| 7. [tex]\( AC = CE \)[/tex] | Substitution property of equality |
Let's break down each part of the proof:
1. Statement: [tex]\( AB = CD \)[/tex]
- Reason: Given
2. Statement: [tex]\( BC = DE \)[/tex]
- Reason: Given
3. Statement: [tex]\( AB + BC = CD + BC \)[/tex]
- Reason: Addition property of equality (we add [tex]\( BC \)[/tex] to both sides of the equalities in previous steps)
4. Statement: [tex]\( CD + BC = CD + DE \)[/tex]
- Reason: Substitution property of equality (replacing [tex]\( BC \)[/tex] with [tex]\( DE \)[/tex] based on step 2)
5. Statement: [tex]\( AB + BC = AC \)[/tex]
- Reason: Segment addition (defining [tex]\( AC \)[/tex] as the combined length of segments [tex]\( AB \)[/tex] and [tex]\( BC \)[/tex])
6. Statement: [tex]\( CD + DE = CE \)[/tex]
- Reason: Segment addition (defining [tex]\( CE \)[/tex] as the combined length of segments [tex]\( CD \)[/tex] and [tex]\( DE \)[/tex])
7. Statement: [tex]\( AC = CE \)[/tex]
- Reason: Substitution property of equality (replacing [tex]\( AB + BC \)[/tex] with [tex]\( AC \)[/tex] and [tex]\( CD + DE \)[/tex] with [tex]\( CE \)[/tex] from steps 5 and 6)
| Statements | Reasons |
|-----------------------------------------------|----------------------------------------------|
| 1. [tex]\( AB = CD \)[/tex] | Given |
| 2. [tex]\( BC = DE \)[/tex] | Given |
| 3. [tex]\( AB + BC = CD + BC \)[/tex] | Addition property of equality |
| 4. [tex]\( CD + BC = CD + DE \)[/tex] | Substitution property of equality |
| 5. [tex]\( AB + BC = AC \)[/tex] | Segment addition (AB + BC = AC) |
| 6. [tex]\( CD + DE = CE \)[/tex] | Segment addition (CD + DE = CE) |
| 7. [tex]\( AC = CE \)[/tex] | Substitution property of equality |
Let's break down each part of the proof:
1. Statement: [tex]\( AB = CD \)[/tex]
- Reason: Given
2. Statement: [tex]\( BC = DE \)[/tex]
- Reason: Given
3. Statement: [tex]\( AB + BC = CD + BC \)[/tex]
- Reason: Addition property of equality (we add [tex]\( BC \)[/tex] to both sides of the equalities in previous steps)
4. Statement: [tex]\( CD + BC = CD + DE \)[/tex]
- Reason: Substitution property of equality (replacing [tex]\( BC \)[/tex] with [tex]\( DE \)[/tex] based on step 2)
5. Statement: [tex]\( AB + BC = AC \)[/tex]
- Reason: Segment addition (defining [tex]\( AC \)[/tex] as the combined length of segments [tex]\( AB \)[/tex] and [tex]\( BC \)[/tex])
6. Statement: [tex]\( CD + DE = CE \)[/tex]
- Reason: Segment addition (defining [tex]\( CE \)[/tex] as the combined length of segments [tex]\( CD \)[/tex] and [tex]\( DE \)[/tex])
7. Statement: [tex]\( AC = CE \)[/tex]
- Reason: Substitution property of equality (replacing [tex]\( AB + BC \)[/tex] with [tex]\( AC \)[/tex] and [tex]\( CD + DE \)[/tex] with [tex]\( CE \)[/tex] from steps 5 and 6)
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