Question

introduction to proofs: justifying statements
qr = st
prove: pq = tu
1 pr = pq + qr reason?
2 su = st + tu reason?
3 pr = su reason?
4 pq + qr = st + tu reason?
5 qr = st reason?
6 pq + st = st + tu reason?
7 pq = tu reason?

Explanation:

Step1: Segment Addition Postulate

The Segment Addition Postulate states that if \(A\), \(B\), and \(C\) are collinear points and \(B\) is between \(A\) and \(C\), then \(AC = AB+BC\). For statement \(1\), since \(Q\) is between \(P\) and \(R\), \(PR = PQ + QR\). For statement \(2\), since \(T\) is between \(S\) and \(U\), \(SU=ST + TU\).

Step2: Given (from the parallel - line and congruent - segment properties)

If two lines are parallel and the segments \(PR\) and \(SU\) are formed by transversals intersecting these parallel lines in a way that the entire lengths of \(PR\) and \(SU\) are equal (by the property of congruent segments formed by parallel lines), \(PR = SU\) (given in the problem - context related to the figure's parallel - line and segment - equality properties).

Step3: Substitution Property of Equality

Since \(PR = PQ + QR\) (from step 1) and \(SU=ST + TU\) (from step 1) and \(PR = SU\) (from step 2), we substitute \(PR\) and \(SU\) in the equations. So, \(PQ + QR=ST + TU\).

Step4: Given

The problem states \(QR = ST\).

Step5: Substitution Property of Equality

Substitute \(QR\) with \(ST\) in the equation \(PQ + QR=ST + TU\). We get \(PQ + ST=ST + TU\).

Step6: Subtraction Property of Equality

Subtract \(ST\) from both sides of the equation \(PQ + ST=ST + TU\). If \(a + b=b + c\), then \(a + b-b=b + c - b\), so \(PQ=TU\).

Answer:

1. Segment Addition Postulate
2. Segment Addition Postulate
3. Given (from figure - related parallel - line and segment - equality properties)
4. Substitution Property of Equality (\(PR = SU\), \(PR = PQ + QR\), \(SU=ST + TU\))
5. Given
6. Substitution Property of Equality (\(QR = ST\))
7. Subtraction Property of Equality