McDougal Littell Jurgensen Geometry: Student Edition Geometry
McDougal Littell Jurgensen Geometry: Student Edition Geometry
5th Edition
ISBN: 9780395977279
Author: Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen
Publisher: Houghton Mifflin Company College Division
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Chapter 3.4, Problem 33WE

a)

To determine

To draw: Parallel and transversal line

a)

Expert Solution
Check Mark

Explanation of Solution

Construction

Two parallel lines AB and CD are drawn and a transversal FE is also drawn.

  McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 3.4, Problem 33WE , additional homework tip  1

b)

To determine

To draw: bisectors of two same sides interior angles.

b)

Expert Solution
Check Mark

Explanation of Solution

Construction

Using protractor, bisectors GI and HI are also drawn

  McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 3.4, Problem 33WE , additional homework tip  2

c)

To determine

To find: the angle form by bisectors.

c)

Expert Solution
Check Mark

Explanation of Solution

On measuring, HIG,we get it to be 90( this is formed by bisectors GI and HI.

  McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 3.4, Problem 33WE , additional homework tip  3

d)

To determine

To show: the angle form by bisectors is 90(.

d)

Expert Solution
Check Mark

Answer to Problem 33WE

The angle formed by the bisectors is 90°

Explanation of Solution

  McDougal Littell Jurgensen Geometry: Student Edition Geometry, Chapter 3.4, Problem 33WE , additional homework tip  4

Proof

   BHG+HGD=180°sum of interior angles on same side of the transversal)

  ⇒ (3+4)+(1+2)=180°

  ⇒ 23+22=180° (since, 1=2and3=4)

  ⇒ 2(3+2)=180°

  ⇒ 2+3=90°…..(i)

  In ΔHIG,

   2+3+5=180°(angle sum property of a triangle)

  ⇒ 90°+5=180°[from (i)]

  ⇒ 5=90°

Chapter 3 Solutions

McDougal Littell Jurgensen Geometry: Student Edition Geometry

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