Question 2

Write the following paragraph proof as a two-column proof.

Given: [tex]\( AB = CD \)[/tex] and [tex]\( BC = DE \)[/tex]
Prove: [tex]\( AC = CE \)[/tex]

Paragraph Proof:

We're given that [tex]\( AB = CD \)[/tex]. By the addition property of equality, we add [tex]\( BC \)[/tex] to both sides of the equation to get [tex]\( AB + BC = CD + BC \)[/tex]. Since we're also given that [tex]\( BC = DE \)[/tex], we use the substitution property of equality to replace [tex]\( BC \)[/tex] with [tex]\( DE \)[/tex] on the right side of the equation. So, [tex]\( AB + BC = CD + DE \)[/tex]. Next, by segment addition, we get that [tex]\( AB + BC \)[/tex] is equal to [tex]\( AC \)[/tex] and that [tex]\( CD + DE \)[/tex] is equal to [tex]\( CE \)[/tex]. Finally, we use the substitution property of equality on the equation [tex]\( AB + BC = CD + DE \)[/tex] to replace [tex]\( AB + BC \)[/tex] with [tex]\( AC \)[/tex] and [tex]\( CD + DE \)[/tex] with [tex]\( CE \)[/tex] to get that [tex]\( AC = CE \)[/tex].

Two-Column Proof:

| Statements | Reasons |
|--------------------------------------------|-------------------------------------|
| 1. [tex]\( AB = CD \)[/tex] | 1. Given |
| 2. [tex]\( BC = DE \)[/tex] | 2. Given |
| 3. [tex]\( AB + BC = CD + BC \)[/tex] | 3. Addition property of equality |
| 4. [tex]\( AB + BC = CD + DE \)[/tex] | 4. Substitution property of equality|
| 5. [tex]\( AB + BC = AC \)[/tex] | 5. Segment addition |
| 6. [tex]\( CD + DE = CE \)[/tex] | 6. Segment addition |
| 7. [tex]\( AC = CE \)[/tex] | 7. Substitution property of equality|

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