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Q12

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Found in: Page 42

### Geometry

Book edition Student Edition
Author(s) Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen
Pages 227 pages
ISBN 9780395977279

# Copy everything shown and write a two-column proof.Given: $RP=TQ;$$PS=QS$Prove: $RS=TS$

 Statements Reasons 1. $\begin{array}{c}RP=TQ;\\ PS=QS\end{array}$ Given 2. $RP+PS=TQ+QS$ 2. Addition property 3. $\begin{array}{c}RS=RP+PS;\\ TS=TQ+QS\end{array}$ 3. Segment addition postulate 4. $RS=TS$ 4. Substitution property
See the step by step solution

## Step 1. Apply the addition property of algebra.

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

$RP+PS=TQ+QS$

## Step 2. Apply segment addition postulate.

From the given figure, it can be observed that sum of $RP$ and $PS$ gives $RS$, that is $RS=RP+PS$.

From the given figure, it can be observed that sum of $TQ$ and $QS$ gives $TS$, that is $TS=TQ+QS$.

## Step 3. Apply Substitution property.

Substitute $RS$ for $RP+PS$ and $TS$ for $TQ+QS$ in step 1.

$\begin{array}{c}RP+PS=TQ+QS\\ RS=TS\end{array}$

Hence, it is proved that $RS=TS$.

## Step 4. Write a two-column proof.

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

 Statements Reasons 1. $\begin{array}{c}RP=TQ;\\ PS=QS\end{array}$ Given 2. $RP+PS=TQ+QS$ 2. Addition property 3. $\begin{array}{c}RS=RP+PS;\\ TS=TQ+QS\end{array}$ 3. Segment addition postulate 4. $RS=TS$ 4. Substitution property