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Recall the Addition and Transitive Properties of Equality.
Statements
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Reasons
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1. PQ=RS
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1. Given
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2. PQ+QR=RS+QR
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2. Addition Property of Equality
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3. PQ+QR=PR
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3. Segment Addition Postulate
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4. RS+QR=QS
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4. Segment Addition Postulate
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5. PR=QS
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5. Transitive Property of Equality
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We will prove the given theorem by filling in the blanks for the given two-column proof. Let's assign a letter to each blank.
Statements
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Reasons
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1. PQ=RS
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1. a.
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2. PQ+QR=QS+QR
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2. b.
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3. c.
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3. Segment Addition Postulate
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4. RS+QR=QS
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4. Segment Addition Postulate
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5. PR=QS
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5. d.
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Let's take a look at the given segment.
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.
Statement1)& PQ = RS Reason1)& a. Given
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
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
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
Finally, we can complete our two-column table!
Statements
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Reasons
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1. PQ=RS
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1. Given
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2. PQ+QR=RS+QR
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2. Addition Property of Equality
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3. PQ+QR=PR
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3. Segment Addition Postulate
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4. RS+QR=QS
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4. Segment Addition Postulate
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5. PR=QS
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5. Transitive Property of Equality
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