Question

1. proof consider \\(\overleftrightarrow{ps}\\).

a. complete the two - column proof.
given: \\(\overline{pq} \cong \overline{rs}\\)
prove: \\(\overline{pr} \cong \overline{qs}\\)
\\(\

$$\begin{array}{|c|c|}\\hline\\text{statement}&\\text{reason}\\\\\\hline1. \\ \\overline{pq} \\cong \\overline{rs}&1. \\ \\underline{\\quad\\quad\\quad\\quad\\quad}\\\\\\hline2. \\ \\underline{\\quad\\quad?\\quad\\quad}&2. \\text{definition of congruent segments}\\\\\\hline3. \\ pq + qr = pr \\text{ and } qr + rs = qs&3. \\ \\underline{\\quad\\quad?\\quad\\quad}\\\\\\hline4. \\ rs + qr = pr&4. \\ \\underline{\\quad\\quad?\\quad\\quad}\\\\\\hline5. \\ qr + rs = pr&5. \\ \\underline{\\quad\\quad?\\quad\\quad}\\\\\\hline6. \\ qs = pr&6. \\ \\underline{\\quad\\quad?\\quad\\quad}\\\\\\hline7. \\ pr = qs&7. \\text{symmetric property of equality}\\\\\\hline8. \\ \\underline{\\quad\\quad?\\quad\\quad}&8. \\text{definition of congruent segments}\\\\\\hline\\end{array}$$

\\)

Explanation:

Step1: Identify Reason for Statement 1

The first statement is given, so the reason is "Given".

Step2: Determine Statement 2

By the definition of congruent segments, if \(\overline{PQ}\cong\overline{RS}\), then their lengths are equal. So \(PQ = RS\).

Step3: Identify Reason for Statement 3

The equations \(PQ + QR=PR\) and \(QR + RS = QS\) come from the Segment Addition Postulate, which states that if a point \(Q\) is between \(P\) and \(R\), then \(PQ+QR = PR\), and similarly for \(Q\), \(R\), \(S\).

Step4: Identify Reason for Statement 4

We substitute \(PQ\) with \(RS\) (from Step 2 where \(PQ = RS\)) in the equation \(PQ+QR = PR\), so this is the Substitution Property of Equality.

Step5: Identify Reason for Statement 5

The equation \(RS + QR=PR\) and \(QR + RS=QS\) (from Step 3) have \(RS + QR\) and \(QR + RS\) which are equal by the Commutative Property of Addition (since \(a + b=b + a\) for real numbers, and lengths are real numbers).

Step6: Identify Reason for Statement 6

We have \(QR + RS=QS\) (from Step 3) and \(QR + RS = PR\) (from Step 5), so by the Transitive Property of Equality (if \(a = b\) and \(b = c\), then \(a = c\)), we get \(QS=PR\).

Step7: Determine Statement 8

By the definition of congruent segments, if \(PR = QS\), then \(\overline{PR}\cong\overline{QS}\).

Filling in the blanks:

1. Reason: Given
2. Statement: \(PQ = RS\)
3. Reason: Segment Addition Postulate
4. Reason: Substitution Property of Equality
5. Reason: Commutative Property of Addition
6. Reason: Transitive Property of Equality
7. Statement: \(\overline{PR}\cong\overline{QS}\)

Answer:

1. Given
2. \(PQ = RS\)
3. Segment Addition Postulate
4. Substitution Property of Equality
5. Commutative Property of Addition
6. Transitive Property of Equality
7. \(\overline{PR}\cong\overline{QS}\)