Answer :
Sure! Here is a two-column proof for the given problem based on the paragraph proof provided:
| Statement | Reason |
|---------------------------------------|-----------------------------------------------|
| 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] and [tex]\( CD + DE = CE \)[/tex] | 5. Segment Addition |
| 6. [tex]\( AC = CE \)[/tex] | 6. Substitution Property of Equality |
Thus, starting from the given statements and following the logical steps, we have shown that [tex]\( AC = CE \)[/tex].
| Statement | Reason |
|---------------------------------------|-----------------------------------------------|
| 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] and [tex]\( CD + DE = CE \)[/tex] | 5. Segment Addition |
| 6. [tex]\( AC = CE \)[/tex] | 6. Substitution Property of Equality |
Thus, starting from the given statements and following the logical steps, we have shown that [tex]\( AC = CE \)[/tex].