Question

1. given: q is between p and r, r is between q and s, pr = qs. prove: pq = rs proof: statements reasons 1. q is between p and r. 1. given 2. pq + qr = pr 2. ______ 3. r is between q and s. 3. ____ 4. ____ 4. seg. add. post. 5. pr = qs 5. ____ 6. pq + qr = qr + rs 6. ____ 7. pq + qr - qr = qr + rs - qr 7. ____ 8. ______ 8. substitution

Explanation:

Step1: Segment Addition Postulate

Since \(Q\) is between \(P\) and \(R\), by the Segment Addition Postulate, \(PQ + QR=PR\). So the reason for statement 2 is "Segment Addition Postulate".

Step2: Given

Statement 3 " \(R\) is between \(Q\) and \(S\)" is given.

Step3: Segment Addition Postulate

By the Segment Addition Postulate, since \(R\) is between \(Q\) and \(S\), \(QR + RS = QS\). So statement 4 is \(QR + RS=QS\).

Step4: Given

Statement 5 "\(PR = QS\)" is given.

Step5: Substitution/Transitive

Substitute \(PR\) with \(PQ + QR\) and \(QS\) with \(QR + RS\) (from statements 2 and 4) into \(PR = QS\). We get \(PQ+QR=QR + RS\). So the reason for statement 6 is "Substitution/Transitive".

Step6: Subtraction Property of Equality

For the equation \(PQ + QR=QR + RS\), subtract \(QR\) from both sides. \(PQ+QR - QR=QR + RS - QR\). The reason for statement 7 is "Subtraction Property of Equality".

Step7: Simplification

Simplify \(PQ+QR - QR=QR + RS - QR\) to get \(PQ = RS\). So statement 8 is \(PQ = RS\).

Answer:

1. Segment Addition Postulate; 3. Given; 4. \(QR + RS=QS\); 5. Given; 6. Substitution/Transitive; 7. Subtraction Property of Equality; 8. \(PQ = RS\)