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Q12

Expert-verified

Geometry

Found in: Page 42

Book edition Student Edition

Author(s) Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen

Pages 227 pages

ISBN 9780395977279

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Short Answer

Copy everything shown and write a two-column proof.
Given: $R P = T Q;$
$P S = Q S$
Prove: $R S = T S$

Statements	Reasons
1. $\begin{matrix} R P = T Q; \\ P S = Q S \end{matrix}$	Given
2. $R P + P S = T Q + Q S$	2. Addition property
3. $\begin{matrix} R S = R P + P S; \\ T S = T Q + Q S \end{matrix}$	3. Segment addition postulate
4. $R S = T S$	4. Substitution property

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Apply the addition property of algebra.

If $a = b$ and $c = d$ , then $a + c = b + d$ .

$R P + P S = T Q + Q S$

Step 2. Apply segment addition postulate.

From the given figure, it can be observed that sum of $R P$ and $P S$ gives $R S$ , that is $R S = R P + P S$ .

From the given figure, it can be observed that sum of $T Q$ and $Q S$ gives $T S$ , that is $T S = T Q + Q S$ .

Step 3. Apply Substitution property.

Substitute $R S$ for $R P + P S$ and $T S$ for $T Q + Q S$ in step 1.

$\begin{matrix} R P + P S = T Q + Q S \\ R S = T S \end{matrix}$

Hence, it is proved that $R S = T S$ .

Step 4. Write a two-column proof.

Write a two-column proof with the help of the above-mentioned steps.

Statements	Reasons
1. $\begin{matrix} R P = T Q; \\ P S = Q S \end{matrix}$	Given
2. $R P + P S = T Q + Q S$	2. Addition property
3. $\begin{matrix} R S = R P + P S; \\ T S = T Q + Q S \end{matrix}$	3. Segment addition postulate
4. $R S = T S$	4. Substitution property

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