Question

complete the proof by choosing the correct
eason\.
given: ( pr = qs )
prove: ( pq = rs )
1 ( pr = pq + qr )
2 ( qs = qr + rs )
3 ( pr = qs )
4 ( pq + qr = qr + rs )
5 ( pq = rs )

Explanation:

Step1: Segment Addition Postulate

The Segment Addition Postulate states that if \(A\), \(B\), and \(C\) are collinear points with \(B\) between \(A\) and \(C\), then \(AC = AB+BC\). For the segment \(PR\) with \(Q\) between \(P\) and \(R\), \(PR = PQ + QR\). Similarly, for the segment \(QS\) with \(R\) between \(Q\) and \(S\), \(QS=QR + RS\). So, the reason for step \(2\) (\(QS = QR+RS\)) is the Segment Addition Postulate.

Step2: Given

We are given in the problem statement that \(PR = QS\). So, the reason for step \(3\) (\(PR = QS\)) is Given.

Step3: Substitution Property of Equality

Since \(PR=PQ + QR\) (step \(1\)), \(QS = QR+RS\) (step \(2\)), and \(PR = QS\) (step \(3\)), we can substitute \(PR\) with \(PQ + QR\) and \(QS\) with \(QR + RS\) in the equation \(PR = QS\). So, \(PQ+QR=QR + RS\). The reason for step \(4\) (\(PQ + QR=QR + RS\)) is the Substitution Property of Equality (if \(a=b\) and \(c = d\) and \(a=c\), then \(b = d\)).

Step4: Subtraction Property of Equality

The Subtraction Property of Equality states that if \(a=b\), then \(a - c=b - c\). In the equation \(PQ+QR=QR + RS\), we can subtract \(QR\) from both sides. \((PQ + QR)-QR=(QR + RS)-QR\). Simplifying, we get \(PQ=RS\). So, the reason for step \(5\) (\(PQ = RS\)) is the Subtraction Property of Equality.

Answer:

1. Segment Addition Postulate
2. Given
3. Substitution Property of Equality
4. Subtraction Property of Equality