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Q5

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Geometry
Found in: Page 174
Geometry

Geometry

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

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Illustration

Short Answer

State the principal definition or theorem that enables you to deduce, from the information given, that quadrilateral SACK is a parallelogram.

∠SKC≅∠CAS; ∠KCA≅∠ASK

The theorem that enables us to deduce from the information given, that quadrilateral SACK is a parallelogram is that if both pairs of opposite angles of the quadrilateral are congruent then the quadrilateral is a parallelogram.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1. Observe the given diagram.

The given diagram is:

Step 2. Description of step.

It is being given that ∠SKC≅∠CAS and ∠KCA≅∠ASK.

That implies it has been given that both pairs of opposite angles of the quadrilateral SACK are congruent.

Step 3. Write the principal definition or theorem that enables you to deduce, from the information given, that quadrilateral SACK is a parallelogram.

The theorem that enables us to deduce from the information given, that quadrilateral SACK is a parallelogram is that if both pairs of opposite angles of the quadrilateral are congruent then the quadrilateral is a parallelogram.

Most popular questions for Math Textbooks

Quadrilateral ABCD is a parallelogram. Name the principal theorem or definition that justifies the statement.

∠ADX≅∠CBX

The quadrilaterals numbered 1, 2, 3, 4, and 5 are parallelograms. If you wanted to show that quadrilateral 6 is also a parallelogram, which of the five methods listed on page 172 would be easiest to use?

Study the markings on each figure and decide whether ABCD must be a parallelogram. If the answer is yes, state the definition or theorem that applies.

Find something interesting to prove. Then prove it. Answers may vary.

Given: ▱EFIH; ▱EGHJ; ∠1≅∠2

Draw and label a diagram. List what is given and what is to be proved. Then write a two-column proof of the theorem.

Theorem 5-5.
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