Question

3
statements:
pq = rs
pq + qr = rs + qr
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rs + qr = qs
pr = qs
:: pq + qr = pr
:: addition property of equality
:: reflexive property of equality
:: substitution property of equality
:: transitive property of equality
reasons:
given
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segment addition postulate (post 1.2)
segment addition postulate (post 1.2)
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buttons: check answer, scientific, calculator

Explanation:

To solve this proof, we analyze each step:

Step 1: Analyze \( PQ = RS \)

• Statement: \( PQ = RS \)
• Reason: Given (as provided in the table)

Step 2: Analyze \( PQ + QR = RS + QR \)

• Statement: \( PQ + QR = RS + QR \)
• Reason: Addition Property of Equality (we add \( QR \) to both sides of \( PQ = RS \))

Step 3: Analyze the middle dashed box (statement)

• Statement: \( PQ + QR = PR \)
• Reason: Segment Addition Postulate (Post 1.2) (this is a given reason - left side is \( PQ + QR \), so by segment addition, it equals \( PR \))

Step 4: Analyze \( RS + QR = QS \)

• Statement: \( RS + QR = QS \)
• Reason: Segment Addition Postulate (Post 1.2) (left side is \( RS + QR \), so by segment addition, it equals \( QS \))

Step 5: Analyze \( PR = QS \)

• Statement: \( PR = QS \)
• Reason: Transitive Property of Equality (since \( PR = PQ + QR \), \( QS = RS + QR \), and \( PQ + QR = RS + QR \), so \( PR = QS \) by transitivity)

Filling in the Table:

Statements	Reasons
\( PQ + QR = RS + QR \)	Addition Property of Equality
\( PQ + QR = PR \)	Segment Addition Postulate (Post 1.2)
\( RS + QR = QS \)	Segment Addition Postulate (Post 1.2)
\( PR = QS \)	Transitive Property of Equality

Final Answer (for the last reason box):

The reason for \( PR = QS \) is the Transitive Property of Equality.

Answer:

To solve this proof, we analyze each step:

Step 1: Analyze \( PQ = RS \)

• Statement: \( PQ = RS \)
• Reason: Given (as provided in the table)

Step 2: Analyze \( PQ + QR = RS + QR \)

• Statement: \( PQ + QR = RS + QR \)
• Reason: Addition Property of Equality (we add \( QR \) to both sides of \( PQ = RS \))

Step 3: Analyze the middle dashed box (statement)

• Statement: \( PQ + QR = PR \)
• Reason: Segment Addition Postulate (Post 1.2) (this is a given reason - left side is \( PQ + QR \), so by segment addition, it equals \( PR \))

Step 4: Analyze \( RS + QR = QS \)

• Statement: \( RS + QR = QS \)
• Reason: Segment Addition Postulate (Post 1.2) (left side is \( RS + QR \), so by segment addition, it equals \( QS \))

Step 5: Analyze \( PR = QS \)

• Statement: \( PR = QS \)
• Reason: Transitive Property of Equality (since \( PR = PQ + QR \), \( QS = RS + QR \), and \( PQ + QR = RS + QR \), so \( PR = QS \) by transitivity)

Filling in the Table:

Statements	Reasons
\( PQ + QR = RS + QR \)	Addition Property of Equality
\( PQ + QR = PR \)	Segment Addition Postulate (Post 1.2)
\( RS + QR = QS \)	Segment Addition Postulate (Post 1.2)
\( PR = QS \)	Transitive Property of Equality

Final Answer (for the last reason box):

The reason for \( PR = QS \) is the Transitive Property of Equality.