Question

pq = rs
pq + qr = rs + qr
segment addition postulate (post. 1.2)
rs + qr = qs
segment addition postulate (post. 1.2)
pr = qs
:: pq + qr = pr :: addition property of equality :: given
:: reflexive property of equality :: substitution property of equality
:: transitive property of equality

Explanation:

Step1: Given equality

$PQ = RS$ is given.

Step2: Apply addition property of equality

Adding $QR$ to both sides of $PQ = RS$ gives $PQ+QR = RS + QR$ by the Addition Property of Equality.

Step3: Use segment addition postulate

By the Segment Addition Postulate, $PQ + QR=PR$ and $RS + QR = QS$.

Step4: Use substitution property of equality

Since $PQ+QR = RS + QR$, $PQ + QR=PR$ and $RS + QR = QS$, we can substitute to get $PR = QS$ by the Substitution Property of Equality.

Answer:

1. Given: $PQ = RS$
2. Addition Property of Equality: $PQ+QR = RS + QR$
3. Segment Addition Postulate: $PQ + QR=PR$
4. Segment Addition Postulate: $RS + QR = QS$
5. Substitution Property of Equality: $PR = QS$