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Q15

Expert-verified

Geometry

Found in: Page 175

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

For exercises, 14-18 write paragraph proofs.
Given: parallelogram ABCD, $\bar{AN}$ bisects $∠ DAB$ ; $\bar{CM}$ bisects $∠ BCD$ .
Prove: AMCN is a parallelogram.

It is proved that the quadrilateral AMCN 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 ABCD is a parallelogram.

In a parallelogram, both pairs of opposite sides are congruent and parallel.

Therefore, in the parallelogram ABCD, both pairs of opposite sides are congruent and parallel and both pairs of opposite angles are congruent.

Therefore, $\bar{A B} ≅ \bar{C D}$ , $\bar{A D} ≅ \bar{B C}$ , $\bar{A B} ∥ \bar{C D}$ and $\bar{A D} ∥ \bar{B C}$ , $∠ B C D ≅ ∠ D A B$ and $∠ A D C ≅ ∠ C B A$ .

Therefore, $A D = B C$ , $\bar{A B} = \bar{C D}$ , $∠ B C D = ∠ D A B$ and $∠ A D C = ∠ C B A$ .

Step 3. Description of step.

It is also being given that $\bar{A N}$ bisects $∠ D A B$ and $\bar{C M}$ bisects $∠ B C D$ .

As, $\bar{A N}$ bisects $∠ D A B$ , therefore by using the definition of angle bisector it can be said that $∠ D A N = \frac{1}{2} ∠ D A B$ .

As, $\bar{C M}$ bisects $∠ B C D$ , therefore by using the definition of angle bisector it can be said that $∠ B C M = \frac{1}{2} ∠ B C D$ .

Therefore, it can be noticed that:

$\begin{matrix} ∠ B C D = ∠ D A B \\ \frac{1}{2} ∠ B C D = \frac{1}{2} ∠ D A B \\ ∠ B C M = ∠ D A N \end{matrix}$

Therefore, $∠ B C M ≅ ∠ D A N$ .

In the triangles $△ D A N$ and $△ B C M$ , it can be noticed that $∠ B C M ≅ ∠ D A N$ , $A D = B C$ and $∠ A D C = ∠ C B A$ .

Therefore, the triangles $△ D A N$ and $△ B C M$ are congruent by ASA congruence.

Therefore, by corresponding parts of congruent triangles, it can be said that $A N ≅ C M$ and $D N ≅ B M$ .

Step 4. Description of step.

As, $\bar{A B} = \bar{C D}$ , therefore it can be obtained that:

$\begin{matrix} \bar{A B} = \bar{C D} \\ \bar{A B} - \bar{B M} = \bar{C D} - \bar{B M} \\ \bar{A B} - \bar{B M} = \bar{C D} - \bar{D N} (∵ \bar{B M} = \bar{D N}) \\ \bar{A M} = \bar{N C} \end{matrix}$

Therefore, it can be noticed that $A N ≅ C M$ and $A M ≅ N C$ .

If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram.

As, $A N ≅ C M$ and $A M ≅ N C$ , therefore, the quadrilateral AMCN is a parallelogram.

Most popular questions for Math Textbooks

Prove Theorem $5 - 1$ .

Write a paragraph proof: The sum of the lengths of the segments drawn from any point in the base of an isosceles triangle perpendicular to the legs is equal to the length of the altitude drawn to one leg.

Prove Theorem 5-3.

You can use a sheet of lined notebook paper to divide a segment into a number of congruent parts. Here a piece of cardboard with edge $\bar{AB}$ is placed so that $\bar{AB}$ is separated into five congruent parts. Explain why it works.

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

$AX = \frac{1}{2} AC$

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