Question

write a paragraph proof for this two - column proof.
given: $overline{ac}congoverline{dc}$ and $overline{ab}congoverline{de}$
prove: $cb = ce$
two - column proof

1. $overline{ac}congoverline{dc},overline{ab}congoverline{de}$ (given)
2. $ac = dc,ab = de$ (def. of $cong$)
3. $ac + cb=ab$ (seg. add. post.)
4. $dc + ce = de$ (seg. add. post.)
5. $ac + ce = de$ (subst. prop. of =)
6. $ac + cb = de$ (subst. prop. of =)
7. $cb = ce$ (trans. prop. of =)

it is given that line segments ac and dc are congruent, as well as line segments ab and de. so, by the definition of congruence, the lengths of line segments ac and ab are equal, as well as dc and de. by the segment addition postulate, ac + cd = ab. it follows that by the substitution segment addition postulate, ab + bc = de. therefore, by the transitive property of equality, cb = ce.
it is given that line segments ac and dc are congruent, as well as line segments ab and de. so, by the definition of congruence, the lengths of line segments ac and dc are equal, as well as ab and de. by the segment addition postulate, ac + cb = ab and dc + ce = de. it follows that cb = de since ab = de. therefore, by the transitive property of equality, cb = ce.
it is given that line segments ac and dc are congruent, as well as line segments ab and de. so, by the definition of congruence, the lengths of line segments ac and de are equal, as well as dc and ab. by the segment addition postulate, ac + cb = ab. it follows that by the substitution segment addition postulate, dc + ce = de. therefore, by the transitive property of equality, cb = ce.

Explanation:

We are given that $\overline{AC} \cong \overline{DC}$ and $\overline{AB} \cong \overline{DE}$. By the definition of congruent segments, their lengths are equal, so $AC = DC$ and $AB = DE$. The Segment Addition Postulate states $AC + CB = AB$ and $DC + CE = DE$. Since $AC = DC$, substitute $AC$ for $DC$ in $DC + CE = DE$ to get $AC + CE = DE$. Because $AB = DE$, substitute $AB$ for $DE$, resulting in $AC + CE = AB$. From $AC + CB = AB$, we equate the two expressions for $AB$: $AC + CB = AC + CE$. Subtracting $AC$ from both sides gives $CB = CE$.

Answer:

It is given that line segments AC and DC are congruent, as well as line segments AB and DE. So, by the definition of congruence, the lengths of line segments AC and DC are equal, as well as AB and DE. By the Segment Addition Postulate, AC + CB = AB and DC + CE = DE. It follows that by substitution, AC + CE = DE (since AC = DC) and AC + CB = DE (since AB = DE). Therefore, by the Transitive Property of Equality, AC + CB = AC + CE, so CB = CE.