Question

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

Explanation:

Step1: Identify given information

Given that $PQ = RS$.

Step2: Apply Segment - Addition Postulate

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

Step3: Use substitution

Since $PQ = RS$, we can substitute $PQ$ for $RS$ in the equation $RS + QR=QS$. So $PQ+QR = QS$.

Step4: Use transitive property

We know $PQ + QR=PR$ and $PQ + QR = QS$. By the transitive property of equality, if $a=b$ and $a = c$, then $b = c$. So $PR=QS$.

Answer:

The proof is completed as above to show that $PR = QS$ using the given $PQ = RS$ and properties like Segment - Addition Postulate, substitution, and transitive property of equality.