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

Given: $\bar{FC}$ and $\bar{SH}$ bisect each other at $A;$ localid="1638250328146" $FC = SH$ .
Prove: $SA = AC$ .

It is proved that $\bar{A C} = \bar{S A}$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Consider the diagram.

Here, $\bar{F C}$ , $\bar{S H}$ bisect each other $A$ and $\bar{F C} = \bar{S H}$

Step 2. State the proof.

As $\bar{F C}$ , $\bar{S H}$ bisect each other $A$ ,

$\bar{A C} = \frac{1}{2} (\bar{F C})$ and $\bar{S A} = \frac{1}{2} (\bar{S H})$

Also,

$\bar{F C} = \bar{S H}$ (Given)

Multiply both sides by $\frac{1}{2}$ .

$\begin{matrix} \frac{1}{2} (\bar{F C}) = \frac{1}{2} (\bar{S H}) \\ \bar{A C} = \bar{S A} \end{matrix}$

Step 3. State the conclusion.

Therefore, $\bar{A C} = \bar{S A}$ (proved).

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Geometry

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Short Answer

Given: $\bar{FC}$ and $\bar{SH}$ bisect each other at $A;$ localid="1638250328146" $FC = SH$ .
Prove: $SA = AC$ .

Step by Step Solution

Step 1. Consider the diagram.

Step 2. State the proof.

Step 3. State the conclusion.

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Given: FC¯ and SH¯ bisect each other at A; localid="1638250328146" FC=SH.Prove: SA=AC.

Step by Step Solution

Step 1. Consider the diagram.

Step 2. State the proof.

Step 3. State the conclusion.

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Given: $\bar{FC}$ and $\bar{SH}$ bisect each other at $A;$ localid="1638250328146" $FC = SH$ .
Prove: $SA = AC$ .