We are told that PQ = RS. This is the mathematical expression of the hypothesis of the theorem. We are asked to prove that PR=QS, which is the conclusion of the theorem.
Blank a.
From the information in the exercise, we are given that PQ and RS are equal.
Statement1)& PQ = RS
Reason1)& a. Given
Blank b.
The equation PQ+QR=RS+QR states that the sum of PQ and QR is equal to the sum of RS and QR. We already know that PQ=RS. By the Addition Property of Equality, we can add QR to both sides of the equation.
PQ+ QR=QS+ QR.
Therefore, we can justify the second step by the Addition Property of Equality.
Statement2)& PQ + QR = RS + QR
Reason2)& b. Addition Property of Equality
Blank c.
In the third step of the proof, we want to use the Segment Addition Postulate. Keep in mind that our goal is to prove that PR=QS. Therefore, we need to describe both PR and QS with lengths that we know are equal. Let's start with PR.
Note that P, Q, and R are collinear. Since Q lies between P and R, we can use the Segment Addition Postulate to describe PR.
PQ + QR = PR
We can complete the third blank with the above equation.
Statement3)& c.PQ + QR = PR
Reason3)& Segment Addition Postulate
Blank d.
The final step of the proof is the result of steps two, three, and four. Note that both the right-hand side of the statement in step two and the left-hand side of the statement in step four are equal to RS+QR. Therefore, we can use the Transitive Property of Equality.
PQ+QR= RS+QR
and
RS+QR=QS
⇓
PQ+QR = QS
We can now use the statement in step three, PQ+QR = PR. Just as we did before, we can use the Transitive Property of Equality.
PQ+QR=PR
and
PQ+QR = QS
⇓
PR = QS
By using Transitive Property of Equality twice and combining previous steps, we prove the final step of the proof.
Statement5)& PR=QS
Reason5)& d.Transitive Property of Equality
