The problem requires completing a geometric proof given the following: $\overline{PQ} \cong \overline{RS}$ and $\overline{QS} \cong \overline{ST}$. The goal is to prove that $\overline{PS} \cong \overline{RT}$. A two-column proof format is provided with some missing statements and reasons.

1. Problem Description

The problem requires completing a geometric proof given the following: $\overline{PQ} \cong \overline{RS}$ and $\overline{QS} \cong \overline{ST}$. The goal is to prove that $\overline{PS} \cong \overline{RT}$. A two-column proof format is provided with some missing statements and reasons.

2. Solution Steps

a. The given information should be stated first: $\overline{PQ} \cong \overline{RS}$, $\overline{QS} \cong \overline{ST}$. The reason is "Given".

b. The congruence of segments can be expressed as the equality of their lengths. So, $PQ = RS$ and $QS = ST$. This is due to the definition of congruence.

c. The length of $\overline{PS}$ is the sum of the lengths of $\overline{PQ}$ and $\overline{QS}$, so $PS = PQ + QS$. Similarly, the length of $\overline{RT}$ is the sum of the lengths of $\overline{RS}$ and $\overline{ST}$, so $RT = RS + ST$. This is by the segment addition postulate.

d. Since $PQ = RS$ and $QS = ST$, by the addition property, $PQ + QS = RS + ST$.

e. Now, using the previous results: $PS = PQ + QS$ and $RT = RS + ST$, and $PQ + QS = RS + ST$. By substitution, $PS = RT$.

f. Finally, if $PS = RT$, then $\overline{PS} \cong \overline{RT}$. The reason is the definition of congruence.

3. Final Answer

Here is the completed proof:

Statements | Reasons

------- | --------

a. $\overline{PQ} \cong \overline{RS}, \overline{QS} \cong \overline{ST}$ | a. Given

b. $PQ = RS, QS = ST$ | b. Definition of Congruence

c. $PS = PQ + QS, RT = RS + ST$ | c. Segment Addition Postulate

d. $PQ + QS = RS + ST$ | d. Addition Property

e. $PS = RT$ | e. Substitution

f. $\overline{PS} \cong \overline{RT}$ | f. Definition of Congruence