Question

1. given: ps = rt, pq = st prove: qs = rs statements 1. ps = rt, pq = st 2. pq + qs = ps 3. st + qs = rt 4. rs + st = rt 5. st + qs = rs + st 6. qs = rs reasons 1. given 2. segment - addition postulate 3. substitution property 4. segment - addition postulate 5. substitution property 6. subtraction property

Explanation:

Step1: Start with given equations

Given \(PS = RT\) and \(PQ=ST\).

Step2: Use segment - addition postulate

We know that \(PQ + QS=PS\) (by the segment - addition postulate which states that if \(Q\) is between \(P\) and \(S\), then \(PQ + QS = PS\)).

Step3: Substitute equal segments

Since \(PQ = ST\), we can substitute \(PQ\) with \(ST\) in \(PQ + QS=PS\), getting \(ST + QS=PS\).

Step4: Use segment - addition postulate again

We know that \(RS+ST = RT\) (by the segment - addition postulate which states that if \(S\) is between \(R\) and \(T\), then \(RS + ST=RT\)).

Step5: Substitute \(PS\) for \(RT\)

Since \(PS = RT\), we can substitute \(RT\) with \(PS\) in \(RS + ST=RT\), getting \(RS + ST=PS\). Then, since \(ST + QS=PS\) and \(RS + ST=PS\), we have \(ST + QS=RS + ST\).

Step6: Subtract \(ST\) from both sides

Subtracting \(ST\) from both sides of the equation \(ST + QS=RS + ST\) (using the subtraction property of equality: if \(a + c=b + c\), then \(a=b\)), we get \(QS = RS\).

Answer:

The proof is shown above to prove \(QS = RS\) starting from the given \(PS = RT\) and \(PQ = ST\) using the segment - addition postulate and substitution and subtraction properties of equality.