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Expert-verified Found in: Page 206 ### Geometry

Book edition Student Edition
Author(s) Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen
Pages 227 pages
ISBN 9780395977279 # Write the reasons that justify the statements. Given: $△ABC\cong △RST$. Prove: localid="1637922009427" $AK>RS$ Statement Proof 1. $△ABC\cong △RST$ Given 2. $\overline{AB}\cong \overline{RS}$, or $AB=RS$ Corresponding parts of congruent triangles are congruent. 3. $AK=AB+BK$ Segment Addition Postulate. 4. $AK>AB$ Property of Inequality If $a=b+c$ and $c>0$ then $c>0$. 5. $AK>RS$ Property of Inequality If $a=b+c$ and $c>0$ then $a>c$.
See the step by step solution

## Step 1. Draw the given diagram and define the congruent property of triangles

Two triangles ABC and RST as shown below are the given congruent triangles. The corresponding parts of the congruent triangles are congruent.

Since $△ABC\cong △RST$ therefore, role="math" localid="1637922313136" $\overline{AB}\cong \overline{RS}$, or $AB=RS$

Hence, the reason for 2nd statement is Corresponding parts of congruent triangles are congruent.

From Segment Addition Postulate. B is a point between A and K. Therefore, $AK=AB+BK$

Hence, the reason for 3rd statement is Segment Addition Postulate.

## Step 2. Property of inequality

If $a=b+c$ and $c>0$ then $a>c$.

Since $AK=AB+BK$ therefore, $AK>AB$.

Hence, the answer to 4 is Property of Inequality If $a=b+c$ and $c>0$ then $a>c$.

## Step 3. Prove statement 5

Consider that

From 3rd statement;

$AK=AB+BK$

From 2nd statement;

$\overline{AB}\cong \overline{RS}$, or $AB=RS$

Therefore, $AK=RS+BK$

Hence from the property of inequality, If $a=b+c$ and $c>0$ then role="math" localid="1637921863353" $a>c$.

Therefore, $AK>RS$

Hence, the answer to 5 is Property of Inequality. ### Want to see more solutions like these? 