Questions: Given: QP TS Prove: QT/TR=PS/SR Statement Reason 1 QP TS 2 ∠RQP ≅ ∠R 3 ∠R ≅ ∠R 4 ΔPQR ∼ ΔSTR 5 QR/TR= /SR Given 2 ∠RQP ≅ ∠RTS 3 ∠R ≅ ∠R 4 . ΔPQR ∼ ΔSTR 5 QR/TR= /SR 6 QR=QT+TR 7 PR=PS+SR 8 (QT+TR)/TR= /SR 9 QT/TR+TR/TR=PS/SR+SR/SR 10 QT/TR+1=PS/SR+1 11 QT/TR= /SR

Given: QP TS Prove: QT/TR=PS/SR Statement Reason 1 QP TS 2 ∠RQP ≅ ∠R 3 ∠R ≅ ∠R 4 ΔPQR ∼ ΔSTR 5 QR/TR= /SR Given 2 ∠RQP ≅ ∠RTS 3 ∠R ≅ ∠R 4 . ΔPQR ∼ ΔSTR 5 QR/TR= /SR 6 QR=QT+TR 7 PR=PS+SR 8 (QT+TR)/TR= /SR 9 QT/TR+TR/TR=PS/SR+SR/SR 10 QT/TR+1=PS/SR+1 11 QT/TR= /SR
Transcript text: Given: $\overline{Q P} \| \overline{T S}$ Prove: $\frac{Q T}{T R}=\frac{P S}{S R}$ Statement Reason $1 \overline{Q P} \| \overline{T S}$ $2 \quad \angle R Q P \cong \angle R$ $3 \angle R \cong \angle R$ $4 \Delta P Q R \sim \triangle S T R$ $5 \frac{Q R}{T R}=\frac{\square}{S R}$ Given $2 \angle R Q P \cong \angle R T S$ $3 \angle R \cong \angle R$ $4 . \triangle P Q R \sim \triangle S T R$ $5 \frac{Q R}{T R}=\frac{\square}{S R}$ $6 Q R=Q T+T R$ $7 P R=P S+S R$ $8 \frac{Q T+T R}{T R}=\frac{\square+\square}{S R}$ $9 \frac{Q T}{T R}+\frac{T R}{T R}=\frac{P S}{S R}+\frac{S R}{S R}$ $10 \frac{Q T}{T R}+1=\frac{P S}{S R}+1$ $11 \frac{Q T}{T R}=\frac{\square}{S R}$
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Solution

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Solution Steps

Step 1: Corresponding Angles Postulate

Since $\overline{QP} \parallel \overline{TS}$, we can say that $\angle RQP \cong \angle RTS$ due to the Corresponding Angles Postulate.

Step 2: Reflexive Property

$\angle R \cong \angle R$ due to the Reflexive Property of Congruence.

Step 3: Angle-Angle Similarity

Since $\angle RQP \cong \angle RTS$ and $\angle R \cong \angle R$, we can conclude that $\triangle PQR \sim \triangle STR$ by the Angle-Angle (AA) Similarity Postulate.

Step 4: Corresponding Sides of Similar Triangles

Because $\triangle PQR \sim \triangle STR$, the ratio of corresponding sides is equal. Therefore, $\frac{QR}{TR} = \frac{PR}{SR}$.

Step 5: Segment Addition Postulate

We can express $QR$ as $QT + TR$ and $PR$ as $PS + SR$ using the Segment Addition Postulate.

Step 6: Substitution

Substituting the expressions from Step 5 into the proportion from Step 4 gives us $\frac{QT + TR}{TR} = \frac{PS + SR}{SR}$.

Step 7: Fraction Algebra

We can separate the fractions on both sides of the equation: $\frac{QT}{TR} + \frac{TR}{TR} = \frac{PS}{SR} + \frac{SR}{SR}$.

Step 8: Simplification

Simplifying the fractions with common numerators and denominators, we get $\frac{QT}{TR} + 1 = \frac{PS}{SR} + 1$.

Step 9: Subtraction Property of Equality

Subtracting 1 from both sides of the equation gives the desired result: $\frac{QT}{TR} = \frac{PS}{SR}$.

Final Answer

\\(\boxed{\frac{QT}{TR} = \frac{PS}{SR}}\\)

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