Geometry: A Complete Course -...

Written by: Larry E. Collins

Geometry:A Complete Course

(with Trigonometry)

Module C – Instructor's Guidewith Detailed Solutions for

Progress Tests

RobbinsCreative

Errata March 2015

Geometry: A Complete Course (with Trigonometry)Module C -Instructor's Guide with Detailed Solutions for Progress Tests

Copyright © 2014 by VideotextInteractive

Send all inquiries to:VideotextInteractiveP.O. Box 19761Indianapolis, IN 46219

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted,in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the priorpermission of the publisher, Printed in the United States of America.

ISBN 1-59676-103-21 2 3 4 5 6 7 8 9 10 - RPInc - 18 17 16 15 14

Letter from the author . . .Good job! I'm proud of you! I'm saying that because, here you are, ready to start Module C of the VideoTextInteractive Geometry Course. And that means you have successfully completed Modules A and B.

So, what do you have to look forward to in this Module? Well, you will be using what you have learned, to“prove” other conditionals. That's right. You are now going to begin “playing the game” of Geometry. And thegreat thing about this game is that everybody wins! Not only are you going to get “really good” at provingstatements, you are going to improve and refine your logical reasoning skills, and your analytical and criticalthinking skills. In other words, you are going to become a much better problem-solver.

Of course, because this will be your first encounter with deductively proving various geometric conditionals,we will go very slowly, making sure you follow the logical reasoning involved. And that means I need to makeyou aware of something very important. This is the pivotal Module in the program. Your level of success indeveloping the ability to logically, and deductively, prove a conditional, will be a major factor contributing toyour success with the rest of the program. Please take it seriously. Review lessons again if you have to, with theaid of the Course Notes. Then tackle the exercises.

That brings us to a very important point. Starting in this unit, you are going to have to make a major shift inthe way you approach the exercises. Very simply, there will now be two types of problems. There will beApplications Exercises, which are more short-answer, and conceptual in nature. Then there will be ProofExercises. And proofs take time. That means, if you try to do all of the exercises in a lesson, you will bespending an inordinate amount of time each day on your Geometry. So we are going back to the way we usedto do exercises. In other words, from now on, we will usually be proving one new theorem in each lesson.Then we will be doing exercises to make sure we understand that theorem. When you get to those Exercises,go ahead and do the first one. Generally, it will ask you to try to prove the new theorem again, on your own.That won't be too difficult, but it will be helpful in reinforcing the new concept. Then, for the moment, don'teven think about doing more proofs. Just focus on doing enough of the Applications Exercises to make sureyou can confidently use that theorem. And even then, you may decide to do just the odd problems. When youfinish with those exercises, if you still have some time left in your normal daily lesson allowance, try one ortwo of the Proof Exercises, and use the Solutions Manual to check. Remember, the Solutions Manual is ateaching tool. So please don't worry if you can't immediately see the statements and reasons in the proofs. Thisis a building process. It will take some time. And, of course, don't hesitate to call us.

With that in mind, let's look at the sequencing in this unit. We will start with two lessons which carefully andclearly show you how to “set up” a proof. In these lessons, we will see clear evidence of the principles of logicwe explored in Module A. More specifically, we will make regular use of the “Law of Syllogism”. If you don'tremember what that is, it might be helpful to quickly review. You will find that concept in Unit I, Part F,Lesson 2. I'm sure it will come back to you right away.

We will then prove specific conditionals (called “theorems”) concerning the various relationships betweengeometric figures. Many of these conditionals will seem quite obvious, and you will probably wonder why wedon't just accept them as additional definitions or postulates. You must remember, however, that our goal is toassume as little as possible, and be able to prove everything else we encounter, as we “play the game”.

I will tell you though, that we will, during the course of this unit, have to accept one more postulate, relative toparallel lines. That's alright. Sometimes, when you play a game, you have to “make up a new rule” when youcome upon a situation which can't be handled using the existing rules. (Continued on next page)

By the way, as an additional help, you will find, at the end of the Student WorkText, a template to use forsetting up proofs (make all the copies you want), and a complete listing of all of the properties, definitions, andpostulates we have developed so far. Further, there is a complete listing of all of the theorems we will beproving in this unit. Feel free to copy these lists, and keep them handy as you work through the exercises,quizzes, and tests in the module. This will actually help with your retention of concepts, especially as youreview for the Unit Test, at the end of the module. Remember, we are not so much interested in your ability tomemorize these statements. We really want to know if you can apply them to situations where you are provingnew relationships.

Well, I think you're ready to begin. I'm sure you will do well, and remember, call us on the help-line, if youhave any difficulty at all.

Thomas E. Clark, Author

Table of ContentsInstructional Aids

Program Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .ivScope and Sequence Rationale . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .vi

Detailed Solutions for Progress TestsUnit III - Fundamental TheoremsPart A - Deductive Proof

LESSON 1 - Direct ProofQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5

LESSON 2 - Indirect ProofQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .9Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .11

Part B - Theorems About Points and LinesLESSON 1 - Theorem 1 - “If a point lies outside a line, then exactly

one plane contains the line and the point.”LESSON 2 - Theorem 2 - “If three different points are on a line,

then at most one is between the other two.”Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .13Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .17

Part C - Theorems About Segments and RaysLESSON 1 - Theorem 3 - “If you have a given ray, then there is exactly one

point at a given distance from the endpoint of the ray.”LESSON 2 - Theorem 4 - “If you have a given line segment, then that

segment has exactly one midpoint.”Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .21Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .23

Part D - Theorems About Two LinesLESSON 1 - Theorem 5 - “If two different lines intersect, then exactly one

plane contains both lines.”LESSON 2 - Theorem 6 - “If in a plane, there is a point on a line, then there

is exactly one perpendicular to the line, through that point.”Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .25Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .29

Module C - Table of Contents i

Part E - Theorems About Angles - Part 1 (one angle)LESSON 1 - Theorem 7 - “If, in a half-plane there is a point on a line, then

there is exactly one other ray through the endpoint of the given ray, such that the angle formed by the two rays has a given measure.”

LESSON 2 - Theorem 8 - “If, in a half-plane, you have an angle, then that angle has exactly one bisector.”Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .33Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .37

Part F - Theorems About Angles - Part 2 (two angles)LESSON 1 - Theorem 9 - “If two adjacent acute angles have their exterior

sides in perpendicular lines, then the two angles are complementary.”LESSON 2 - Theorem 10 - “If the exterior sides of two adjacent angles are

opposite rays, then the angles are supplementary.”Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .41Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .45

LESSON 3 - Theorem 11 - “If you have right angles, then those right angles are congruent.”

LESSON 4 - Theorem 12 - “If you have straight angles, then those straight angles are congruent.”Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .51Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .57

Part G - Theorems About Angles - Part 3 (more than two angles)LESSON 1 - Theorem 13 - “If two angles are complementary to the same

angle or congruent angles, then they are congruent to each other.” LESSON 2 - Theorem 14 - “If two angles are supplementary to the same angle

or congruent angles, then they are congruent to each other.” LESSON 3 - Theorem 15 - “If two lines intersect, then the vertical angles

formed are congruent.” Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .63Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .67

ii Module C - Table of Contents

Part H - Theorems About Parallel LinesLESSON 1 - Postulate 11- Corresponding Angles of Parallel Lines LESSON 2 - Theorem 16 - “If two parallel lines are cut by a transversal,

then alternate interior angles are congruent.”LESSON 3 - Theorem 17 - “If two parallel lines are cut by a transversal,

then interior angles on the same side of the transversal are supplementary.” Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .71Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .77

LESSON 4 - Theorem 18 - “If a given line is perpendicular to one of two parallel lines, then it is perpendicular to the other.”

LESSON 5 - Theorem 19 - “If two lines are cut by a transversal so that corresponding angles are congruent, then the two lines are parallel.”

LESSON 6 - Theorem 20 - “If two lines are cut by a transversal so that alternate interior angles are congruent, then the two lines are parallel.”

LESSON 7 - Theorem 21 - “If two lines are cut by a transversal so that interior angles on the same side of the transversal are supplementary,then the two lines are parallel.”

Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .83Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .89

LESSON 8 - Theorem 22 - “If two lines are perpendicular to a third line,then the two lines are parallel.”

LESSON 9 - Theorem 23 - “If two lines are parallel to a third line, then thetwo lines are parallel to each other.”

LESSON 10 - Theorem 24 - “If two parallel planes are cut by a third plane, then the two lines of intersection are parallel.” Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .95Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .99

Unit III Test - Form A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

Unit III Test - Form B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

Module C - Table of Contents iii

iv

Program OverviewThe VideoTextInteractive Geometry program addresses two of the most important aspects ofmathematics instruction. First, the inquiry-based video format contributes to the engaging ofstudents more personally in the concept development process. Through the frequent use of the pausebutton, you, as the instructor, can virtually require interaction and dialogue on the part of your student.As well, students who work on their own, can “simulate” having an instructor present by pausing thelesson every time a question is asked, and trying to answer it correctly before continuing. Of course, thestudent may answer incorrectly, but the narrator will be sure to give the right answer when the playbutton is pressed to resume the lesson. Right or wrong, however, the student is regularly engaging inanalytical and critical thinking, and that is a healthy exercise, in and of itself. Second, eachincremental concept is explored in detail, using no shortcuts, tricks, rules, or formulas, and no stepin the process is ignored. As such, the logic and the continuity of the development assure students thatthey understand completely. Subsequently, learning is more efficient, and all of the required concepts(topics) of the subject can be covered with mastery. Of course, the benefits of these efforts can be seeneven more clearly in a description of a typical session, as follows:

After a brief 2 or 3 sentence introduction of the concept to be considered, usually by examining thedescription, and the objective given at the beginning of the video lesson, you and your student can begin.You should pause the lesson frequently, usually every 15-20 seconds (or more often if appropriate), toengage your student in discussion. This means that, for a 5-10 minute VideoText lesson, it may take 10-15 minutes to finish developing the concept. Dialogue is a cornerstone. In addition, during this time,your student should probably not be allowed to take notes. Students should not have their attentiondivided, or they risk missing important links. Neither should you be dividing your attention, by lookingat notes, or writing on a pad, or an overhead projector. Everyone should be concentrating on conceptdevelopment and understanding. Please understand that a student who is accustomed to workingalone, or can be motivated to study independently, has, with the VideoText, a powerful resource toexplore and master mathematical concepts by simulating the dialogue normally encountered with a“live” instructor. And, because of the extensive detail of the explanations, along with the computergenerated graphics, and animation, students are never shortchanged when it comes to the insightnecessary to fully comprehend.

Once the concept is developed, and the VideoText lesson is completed, you can then employ the CourseNotes to review, reinforce, or to check on your student's comprehension. These Course Notes arereplications of the essential content that was viewed in the VideoText lesson, illustrating the same terms,diagrams, problems, numbers, and logical sequences. In fact, at this time, if your student needs a littlemore help, he or she can use the Course Notes while viewing the lesson again, using them as a guide, tore-examine the concept. The key here is that students concentrate on understanding first, and takecare of documentation later.

Please understand that it is not the intent of the program to let the VideoText lesson completely take theplace of personal instruction or interaction. Actually, the video should never tell your studentsanything that hasn't been considered or discussed (while the lesson is paused), and it should neveranswer questions that have not already been considered and resolved. As such, it becomes a “newbreed” of chalkboard or overhead projector, whereby you, as the teacher, or your student working alone,can “write”, simply by pressing the “play” button. This is a critical point to be understood, and should

v

serve to help you examine all of the materials and strategies from the proper perspective.Next, your student can begin to do some work independently, either by your personal introduction ofadditional examples from the WorkText, or by the student immediately going to the WorkText on his orher own. The primary feature of the WorkText, beside providing problem banks with which studentscan work on mastery, is that objectives are restated, important terms are reviewed, and additionalexamples are considered, in noticeable detail, taking students, once again, through the logic of theconcept development process. The premise here is simple. When students work with an instructor,whether doing exercises on their own, or working through them with other students, they are usuallyconcentrating more on “how to do” the problems. Then, when they leave the instructor, they simplydon't take the discussion of the concept with them. The goal of this program is to provide a resourcewhich will help students “re-live” the concept development on their own, whether for review, or foradditional help. That is the focus of the Student WorkText.

Having completed the exercises for the lesson being considered, your student is now ready to use thedetail in the Solutions Manual to check work and engage in error analysis. Again, it is essential to astudent's understanding that he or she find mistakes, correct them, and be required to give someexplanation, either verbal or in writing, to you as the instructor. In fact, at this stage, you might evenconsider grading your student only on the completion of the work, not on its accuracy. Remember,this is the first time the student has tried to demonstrate understanding of a concept, and he or she maystill need some fine-tuning. So, because this is part of the initial learning process, the focus should beon a careful analysis of the logic behind the work, not just the answers. Finally, it is time to assessyour student's mastery of the concept behind the work. Just be sure you are not testing on the sameday the exercises were completed. Short-term memory can trick you into thinking that you “have it”,when, in fact, you are just remembering what you did moments before. A more accurate evaluation canbe made on the next day, before moving on to the next lesson. Further, the quizzes and tests in theprogram often utilize open-response questions which will require your student to state, in writing,his or her understanding of the concept. This often reveals much more about a student'sunderstanding than just checking to see if an answer on a test is correct. Remember too, that there aretwo versions of every quiz and test, allowing you to retest, if necessary, in order to make sure that yourstudent has mastered the concept.

Of course, just as with the WorkText, there are detailed solutions for all of the quiz and test problems,in the Instructor's Guide. Again, your student should be required to analyze problems that weremissed, and explain why the problem should have been done differently. It is simply a fact that one ofthe most powerful and effective teaching tools you can employ, is to ask your students to “articulate”to you what their thinking was, as they worked toward a given answer.

As you can see, the highly interactive quality of this program, affords students a much greateropportunity than usual to grow mathematically, at a personal level, and develop confidence in theirability. That can have a tremendous impact on a student's future pursuits, especially in an age whereapplications of mathematics are so important.

vi

Scope and Sequence RationaleThere are two basic premises which drive concept development in Geometry, and these two essentialsshape the logical scope and sequence of geometric content.

First, it is generally understood that Geometry is the study of spatial relations. In the same way thatAlgebra is the study of numerical relations (equations and inequalities), and Calculus is concernedprimarily with rates of change, Geometry is a comprehensive exploration of “shapes” (as sets of points),the measurements associated with those shapes, and the relationships that can be established betweenthose shapes. As such, no treatment of Geometry should ever investigate those relationships onlyindividually, or in isolation. This is especially noticeable with traditional textbooks, which generally usea format which addresses them in different “chapters”. In the VideoText Interactive Geometry course,concepts are discussed from a “Unit” perspective, pursuing and connecting, in an exhaustive way,all of the outcomes associated with various possibilities for a specific relationship. Of course, asmuch as is possible, students need to “see” those relationships, and experience the “motion”, or“transformation”, necessary to clearly illustrate the concept. It really is impossible to put a value on thebenefits of visualization, in life in general, and in Geometry in particular. So, in the VideoTextInteractive Geometry program, computer-generated graphics are used extensively, along withanimation and color-sequencing, in order that students can actually see the relationships develop.

The second premise is that geometric concepts should be studied utilizing all of the power andconviction that both inductive and deductive reasoning can bring to the table. In other words, it isalways desirable, and helpful, for students to “experiment”, inductively, with a geometric relationship, inan effort to come to some general conclusion. Once that general conclusion has been arrived at,however, it is even more convincing if the student is able to “prove”, deductively, that the conclusionabsolutely must follow, logically, from the given information. No, formal proof is not often asked for ineveryday life. On the other hand, the exercise of developing that kind of thinking is invaluable, not onlyin some specific job-related activities, but, more generally, in the daily problem-solving situations thatconfront us. The VideoText Interactive Geometry program is formatted in such a way that formal proofis a cornerstone.

Unit I, then, focuses on a complete preparation for students to begin a formal study of Geometry by“re-teaching” of all of the basic geometric concepts for which students have simply memorized theappropriate term, definition, or formula. That means we must re-establish that Mathematics in general,and Geometry in particular, is a language, with parts of speech and sentence structure. We mustdevelop, in detail, the concepts associated with building geometric shapes. We must investigate, againin detail, the concepts dealing with the measurement of those shapes. Finally, we must thoroughlydevelop the principles of inductive and deductive reasoning, giving significant attention to the dynamicsof mathematical deductive logic, which are the building blocks that students will use to construct formalproofs.

In Unit II, we begin the actual study of “Plane Geometry” by developing all of the necessary terms,definitions, and assumptions we will be using as a basis for studying geometric relationships. Inother words, we draw on the analogy that studying any area of Mathematics is like “playing a game”.We must first determine which basic elements will be “undefined” in our Geometry, or accepted

vii

without definition. We must then determine which basic elements can be formally defined, using thoseundefined terms. Finally, we must build a list of “postulates”, or conditional assumptions which willserve as the “rules of the game”, guiding us through the investigation of relationships, in our Geometry.It is important to note, at this point, that every Plane Geometry study will, in certain ways, be unique tothe philosophy of the instructor, depending on the acceptance of these fundamental terms. In otherwords, while the prevailing context will always be that of classical Euclidean Geometry, the lists ofdefinitions and postulates may differ from person to person. The key, however, is that each study willrely on its own particular list of Essential Elements to prove the rest of the relationships to beinvestigated.

So, in Unit III, we use the Fundamental Terms developed in Unit II, to prove FundamentalTheorems related to points, lines, rays, segments, and angles. These theorems will be foundational tothe study of Simple Closed Plane Curves, which are the primary backdrop of all studies of PlaneGeometry.

At this point, since we have put in place the “rules of the game”, we can begin, and, for all practicalpurposes, complete, a methodic investigation of the geometric relationships associated with Triangles(Unit IV), Other Polygons (Unit V), and Circles (Unit VI). That then allows us to conclude our studyby the investigation of several applications, internal to the study of Geometry.

First, in Unit VII, we will engage in the classic geometric exploration of “Construction”. Thismeans that, with the use of only a straight edge (to construct lines, rays, and segments), and a compass(to construct circles, and arcs of circles), we will attempt to use our knowledge of geometric relationshipsto “build”, and “operate on”, various geometric shapes. Included will be the replication and division ofline segments and angles, the building of polygons to desired specifications, and the generation of circlesto desired specifications.

Second, in Unit VIII, we will examine, in significant detail, the relationships between the variouscomponents of triangles. This is, of course, the study of Trigonometry, from the Greek, meaning “tri-angle-measure”. Included are the basic relationships of sine, cosine, and tangent, as well asapplications involving the Pythagorean Theorem, the Laws of Sines and Cosines, and several otherambiguous cases.

Please understand that the organizational argument presented here is not meant to stifle the creativity of the instructor. Neither should it prohibit the instructor from utilizing a modular approach to conceptdevelopment. It does, however, serve to remedy the fragmented, isolated topic, “chapter” approach to a subject which has been traditionally presented to us in “textbooks”, without that element ofdevelopmental continuity. To that end, it speaks loudly to the curricular issues which all instructors face, and the attitudinal issues students deal with when they are presented with a new and differentMathematics course.

Name

Class Date Score

Quiz Form A

Unit III - Fundamental TheoremsPart A - Deductive ProofLesson 1 - Direct Proof

1. Complete the deductive proof below by supplying the missing reasons.

© 2014 VideoTextInteractive Geometry: A Complete Course 1

Conditional: If Diagram: —Not applicable—

Given: Prove:

1. 1. Given

2. 2. Multiplication for Equality

3. 3. Arithmetic Fact

4. 4. Associativity for •


6. 6. Multiplicative Inverse

7. 7. Addition for Equality

8. 8. Distribution of • over +


10. 10. Additive Inverse

11. 11. Multiplication by zero

12. 12. Addition of zero

13. 13. Multiplication of Equality


15. 15. Associativity for •


17. 17. Multiplication by 1

3x = 6 -12

x

2 3x = 2 6 - 212

x

2 3x = 12 - 212

x

( ) ( )

( )

⋅( ) ⋅

⋅

2 3 x = 12 - x

6x = 12 -

212

212

xx

6x = 12 - 1x

6x + 1x = 12 - 1x + 1x

6 + 1 x = 12+ -1+ 1 x

7x

( ) ( )== 12+ -1+ 1 x

7x = 12+ 0 x

7x = 12+ 0

7x = 12

17

7x =17

1

( )⋅

( ) 22

17

7x =127

17

7 x =127

1 x =127

x =127

( )

( )

⋅

⋅

3x = 6 -12

x, then x =127

3x = 6 -12

x x =127

STATEMENT REASON

Name

Class Date Score

Quiz Form B

Unit III - Fundamental TheoremsPart A - Deductive ProofLesson 1 - Direct Proof


1. Complete the deductive proof below by supplying the missing reasons.

Conditional: If , then Diagram: —Not applicable—

Given: Prove:

1. 1. Given

2. 2. Multiplication of Equality

3. 3. Distributivity of • over +




7. 7. Multiplicative Identity

8. 8. Addition of Equality

9. 9. Commutativity of +




13. 13. Multiplication by 0

14. 14. Additive Identity

15. 15. Addition for Equality


17. 17. Additive Identity


19. 19. Multiplication for Equality


21. 21. Associativity of •


23. 23. Multiplicative Identity

x - 2 =2x + 8

5

5 x - 2 = 52x + 8

5

5 x - 5 2 = 52x + 8

5

( )

⋅ ⋅

⋅ ⋅ ⋅ ( )⋅ ⋅ ⋅

5 x - 5 2 = 1 2x + 8

5x - 5 2 = 1 2x + 1 8

5x - =10 11 2x + 1 8

5x - 10 = 2x + 8

5x - 10 - 2x = 2x + 8 - 2x

5x - 2x - 10

⋅ ⋅

== 2x - 2x + 8

5 - 2 x - 10 = 2 - 2 x + 8

3x - 10 = 2 - 2 x + 8

3x

( ) ( )( )

-- 10 = 0x + 8

3x - 10 = 0 + 8

3x - 10 = 8

3x - 10 + 10 = 8 + 10

3x + 0 = 88 + 10

3x = 8 + 10

3x = 18

13

3x = 18

13

3x =

13

3

( ) ( )

( )

⋅

13

6

⋅

⋅

x = 6

1 x = 6

x = 6

xx− = +

22 8

5x = 6

xx− = +

22 8

5x = 6

STATEMENT REASON

Name

Class Date Score

Quiz Form A

Unit III - Fundamental TheoremsPart A - Deductive ProofLesson 2 - InDirect Proof

1. State the negation of each statement

a) It will not rain. ___________________________________________________________________

b) nABC is an isosceles triangle. ______________________________________________________

c) “x + 5” is not an open phrase. _______________________________________________________

2. Indicate whether each pair of statements would enable you to arrive at a contradiction in an indirect proof, and give some justification for your answer.

a) AB < 15; AB > 20 ________________________________________________________________

________________________________________________________________________________

b) /X and /Y are obtuse angles; /X and /Y are supplementary. ___________________________

________________________________________________________________________________

c) Point B is between points A and C; Points A, B, and C are not collinear. _____________________

________________________________________________________________________________

d) /P and /Q are congruent; /P and /Q are complementary. ______________________________

________________________________________________________________________________

3. For each of the following conditionals, state the assumption you would use to start an indirect proof.

a) If a triangle is equilateral, then the triangle is isosceles. __________________________________

_______________________________________________________________________________

b) If an angle is a right angle, then the angle is equal to its supplement. _______________________

_______________________________________________________________________________


It is not true that it will not rain. Or, it will rain.

It is not true that nABC is an isoceles triangle. Or, nABC is not an isoceles triangle.

It is not true that “x + 5” is not an open phrase. Or “x + 5” is an open phrase.

Yes. These two statements are already contradictory

Yes. /X and /Y are greater than 90.

Yes, by the definition

No. Both can be true without contradiction

Suppose (assume) the triangle is not isoceles.

Suppose (assume) the angle is

not equal to its supplement.

of Betweenness

Name

Class Date Score

Quiz Form B

Unit III - Fundamental TheoremsPart A - Deductive ProofLesson 2 - InDirect Proof

1. State the negation of each statement

a) The adjacent sides are not parallel. ________________________________________________________

b) , n. ______________________________________________________________________________

c) “5 – 3 = 7” is a closed phrase. ___________________________________________________________

2. Indicate whether each pair of statements would enable you to arrive at a contradiction in an indirect proof, and give some justification for your answer.

a) , || m; Point X is contained in , and m. ___________________________________________________

___________________________________________________________________________________

b) /A and /B form a linear pair; m/A < 90 and m/B < 90____________________________________

___________________________________________________________________________________

c) /M and /N are vertical angles; /M and /N are obtuse angles. ______________________________

___________________________________________________________________________________

d) Point D is the midpoint of AC; Points A, D, and C are not collinear. ____________________________

___________________________________________________________________________________

3. For each of the following conditionals, state the assumption you would use to start an indirect proof.

a) If two adjacent angles are supplementary, then the angle bisectors of the two angles

are perpendicular. ____________________________________________________________________

___________________________________________________________________________________

b) If two adjacent angles are supplementary, then their exterior sides lie in a straight line.

___________________________________________________________________________________

___________________________________________________________________________________


It is not true that the adjacent sides are not parallel. Or, the adjacent sides are parallel.

It is not true that , n. Or, , is not perpendicular to n.

It is not true that “5 – 3 = 7” is a closed phrase. Or, 5 – 3 = 7 is not a closed phrase.

Yes, parallel lines have no points in common.

Yes, linear pairs of angles have measures which

No, both could be true without contradiction.

Yes, by the definition of a Midpoint.

Suppose (assume) the angle bisectors of the two supplementary adjacent angles are

Suppose the exterior sides of the two supplementary adjacent angles do not lie in a straight line.

> >

add to 180.

not perpendicular.


Name

Class Date Score

Quiz Form B

Unit III - Fundamental TheoremsPart B - Theorems About Points and LinesLesson 1 - Theorem 1: “If a point lies outside a line, then exactly one plane contains the line and the point.”Lesson 2 - Theorem 2: “If three different points are on a line, then at most one is between the other two.”

1. Referring to the diagram at the right, findthe length of PR, if PR > QS, RS = 6, and QR = 4.

Answer__________________________________

2. Given: IK > EA as shown

Prove: IE > KA

1. IK > EA 1. Given

2. IK = EA 2. Definition of Congruent Segments

3. KE = KE 3. Reflexivity for Equality

4. IK + KE = EA +KE 4. Addition for Equality

5. IK + KE = KE + EA 5. Commutativity of Addition

6. IK +KE = IE 6. Postulate 6 (Ruler) - Segment-Addition Assumption

7. KE + EA = KA 7. Postulate 6 (Ruler) - Segment-Addition Assumption

8. IE = KA 8. Substitution (from statements 5, 6, and 7)

9. IE > KA 9. Definition of Congruent Segments

QR + RS = QS

4 + 6 = QS

10 = QS

PR > QS

PR = QS

PR = 10

P S TQ R

J A

I N

K

E P

R

M

K P NB

X Z

P S TQ R

P S TQ R

J A

I N

K

E

S

P

R

ZQ

T

M

K

J

K P NB

X Z

P S TQ R

STATEMENT REASON

PR = 10

NameUnit III, Part B, Lessons 1&2, Quiz Form B—Continued—

© 2014 VideoTextInteractive Geometry: A Complete Course20

5. Answer the following questions regarding the two triangles shown in problem 2.

a) Are the two triangles necessarily in the same plane?

b) Why or why not?

c) Does the relationship of the two triangles have any effect on the proof in problem 2?

d) Why or why not?

No

Theorem 1 states “If a point lies outside a line, then exactly one plane contains the line and the point.” The

triangle formed by Point J and the two points A and K on AK determine exactly one plane. The triangle formed by

point N, I and E on IE determine exactly one plane. AK and IE are the same line since points A, E, K, and I are

collinear. However, point J and point N do not have to be in the same plane. The triangles would then be in

different planes. (Remember, an infinite number of planes can contain one line, and plane JAK and plane NIE

would be two of those planes.)

No

The focus of the problem is the line IA. So, it really doesn’t matter where the rest of the points and line segments

are, that make up the triangles.

J A

I N

K

E

X Z

Name

Class Date Score

Quiz Form A

Unit III - Fundamental TheoremsPart C - Theorems About Segments and RaysLesson 1 - Theorem 3: “If you have a given ray, then there is exactly one point at a given distance from the endpoint of the ray.”Lesson 2 - Theorem 4: “If you have a given line segment, then that segment has exactly one midpoint.”

1. If AB > CD, is there a point X on AB such that AX = CD? Why or Why not?

2. PQ and PR are opposite rays. The coordinate of P is zero. The coordinate of R is 12. Is the coordinate of Q positive or negative? Why or why n not?

3. Draw a diagram that illustrates the information given in the following problems.a) Point C is on AB, AC = 5, and AB is 8.

b) AB and CD intersect, but they are not collinear.

4. In the following problems, tell whether AB and AC are opposite rays. Answer “yes”, “no” or “not enough information”. Then draw a simple diagram to illustrate your answer.a) AB = 6 and AC = 3.

b) B is the midpoint of AC.


Yes. Intuitively if you

The coordinate of Q must be negative. Since PR and PQ go in

opposite directions, starting at 0, Q must be on the negative side of 0.

Not enough information

No

N P

n p,

Q P R

0 12

0 5 8

CA B

Z A B

a0 b

Y QX Z

B A C

M T N

O S

C

DA B

N P

n p,

Q P R

0 12

0 5 8

CA B

Z A B

a0 b

Y QX Z

M T N

O S

C

DA B

“translated” CD onto AB so that C is coincident with A, D would be between A and B and you could rename D as X,

making AX = CD

Answers will vary. One possibility is shown


One possibility is: Another possibility is:

N P

n p,

Q P R

0 12

0 5 8

CA B

Z A B

a0 b

Y QX Z

B A C

M T N

O S

C

DA B

BA C

BAC

CA B

CB A

BAC

-2 1 4

N P

n p,

Q P R

0 12

0 5 8

CA B

Z A B

a0 b

Y QX Z

B A C

M T N

O S

C

DA B

BA C

BAC

CA B

CB A

BAC

CA B

-2 1 4

N P

n p,

Q P R

0 12

0 5 8

CA B

Z A B

a0 b

Y QX Z

B A C

M T N

O S

C

DA B

BA C

BAC

CA B

CB A

BAC

2 1 4

Name

Class Date Score

Quiz Form B


Unit III - Fundamental TheoremsPart C - Theorems About Segments and RaysLesson 1 - Theorem 3: “If you have a given ray, then there is exactly one point at a given distance from the endpoint of the ray.”Lesson 2 - Theorem 4: “If you have a given line segment, then that segment has exactly one midpoint.”

1. On AB, if point X has coordinate 8, can a different point Y on AB have the coordinate 8? Why or why not?

2. On MN, the coordinate of M is zero, and the coordinate of N is 5. If the coordinate of T is positive, is T on MN? Why or why not?

3. Draw a diagram that illustrates the information given in the following problems.a) Points X, Y, Q and Z are collinear. XY and QZ do not intersect.

b) The intersection of two rays, BC and AC, is a segment, AB.

4. In the following problems, tell whether AB and AC are opposite rays. Answer “yes”, “no” or “not enough information”. Then draw a simple diagram to illustrate your answer.a) A is midpoint of BC.

b) The positive real numbers are paired with points on AB.

No, theorem 3 clearly states that exactly one point will be at a given distance from the endpoint of the ray.

The coordinate of T is positive. For T to be negative, the point would have to be on an

opposite ray.

N P

n p,

Q P R

0 12

0 5 8

CA B

Z A B

a0 b

Y QX Z

B A C

M T N

O S

C

DA B

BA C

BAC

CA B

CB A

BAC

CA B

-2 1 4

1 3 5

N P

n p,

Q P R

0 12

0 5 8

CA B

Z A B

a0 b

Y QX Z

B A C

M T N

O S

C

DA B

BA C

BAC

CA B

CB A

BAC

CA B

-2 1 4

1 3 5



One possibility is:

One possibility is: Another possibility is:

yes

Not enough information

N P

n p,

Q P R

0 12

0 5 8

CA B

Z A B

a0 b

Y QX Z

B A C

M T N

O S

C

DA B

BA C

BAC

CA B

CB A

BAC

CA B

-2 1 4

1 3 5

N P

n p,

Q P R

0 12

0 5 8

CA B

Z A B

a0 b

Y QX Z

B A C

M T N

O S

C

DA B

BA C

BAC

CA B

CB A

BAC

CA B

-2 1 4

1 3 5

N P

n p,

Q P R

0 12

0 5 8

CA B

Z A B

a0 b

Y QX Z

B A C

M T N

O S

C

DA B

BA C

BAC

CA B

CB A

BAC

CA B

-2 1 4

1 3 5


Name

Class Date Score

Quiz Form A

Unit III - Fundamental TheoremsPart D - Theorems About Two LinesLesson 1 - Theorem 5: “If two different lines intersect, then exactlyone plane contains both lines.”Lesson 2 - Theorem 6: “If in a plane, there is a point on a line, thenthere is exactly one point perpendicular to the line, through that point.”

1. In the diagram to the right, BE AC and BD BF. Find the measure of each of the following angles.

a) m/EBF ________ b) m/DBE ________

c) m/DBA ________ d) m/DBC ________

2. In the diagram to the right, BE AC and BD BF. Also assume m/CBF = x. Express the measure of each of the following angles:

a) m/EBF ________ b) m/DBE ________

c) m/DBA ________ d) m/DBC ________

3. In the diagram to the right, BE AC and BD BF. Find the value of x in each of the following problems.a) m/DBE = 3x, m/EBF = 4x–1 x = ____________

b) m/ABD = 6x, m/DBE = 3x + 9, x = ____________

m/EBF = 4x + 18, m/FBC = 4x

> >

A B C

,

DF

E

35O

FE

M

RS

A

H G

y

P

C

D

BA

A B C

,

DF

E

35O

A B C

DF

E

FE

M

RS

A O E

D

B

H

C

G

x

y A B

DE

z

90 – 35 = 55 90 – 55 = 35

90 – 35 = 55 90 + 35 = 125

90 – (90 – x) or x

3x + (4x – 1) = 907x – 1 = 90

7x = 91x = 13

6x + 3x + 9 + 4x + 18 + 4x = 18017x + 27 = 180

17x = 153x = 9

55O 35O

55O 125O

90 – x x

90 + x90 – x

A B C

,

DF

E

35O

A B C

DF

E

FE

M

RS

A O E

D

B

F

H

C

G

x

y A B

DE

P

C

D

BA

DB

E

CA

12 4

3

z

(x)O

(3x)O13

9

> >

> >

(4x – 1)O


Unit III, Part D, Lessons 1&2, Quiz Form A—Continued—

Name

4. Using the format given below, write a complete proof with the

“Given” and “Prove” information, and the “Diagram”.

Given: m/1 = m/2 and Diagram:

m/3 = m/4 as shown

Prove: YS XZ

1. m/1 = m/2 1. Given

2. m/3 = m/4 2. Given

3. m/1 + m/3 = m/2 + m/4 3. Addition for Equality

4. m/1 + m/3 = m/XYS 4. Postulate 7 (Protractor) - Angle-Addition Assumption

5. m/2 + m/4 = m/SYZ 5. Postulate 7 (Protractor) - Angle-Addition Assumption

6. m/XYS = m/SYZ 6. Substitution (from statements 3, 4, and 5)

7. /XYZ is a straight angle whose 7. Definition of Straight Anglemeasure of 180O

8. m/XYS + m/SYZ = m/XYZ 8. Postulate 7 (Protractor) - Angle-Addition Assumption

9. m/XYS + m/SYZ = 180 9. Substitution (from statements 7 and 8)

10. m/XYS + m/XYS = 180 10. Substitution (from statements 6 and 9)

11. 2m/XYS = 180 11. Collect like terms (Distributivity)

12. m/XYS = 90 12. Multiplication for Equality

13. /XYS is a right / 13. Definition of Right Angle

14. YS XZ 14. Definition of Perpendicular Lines

C

F

E

35O

A B C

DF

E

X Y Z

S TR

3 4

1 2

F

R

A O E

D

B

F

H

C

G

x

y A B C

DF

E

B

DB

E

CA

12 4

3

z

>

>

STATEMENT REASON


Name

Class Date Score

Quiz Form B

Unit III - Fundamental TheoremsPart D - Theorems About Two LinesLesson 1 - Theorem 5: “If two different lines intersect, then exactlyone plane contains both lines.”Lesson 2 - Theorem 6: “If in a plane, there is a point on a line, thenthere is exactly one point perpendicular to the line, through that point.”

1. In the diagram to the right, BF AE , m/BOC = x, and m/GOH = y. Express the measure each of the following angles in terms of x, y, or both.

a) m/COA ________ b) m/COH ________

c) m/HOF ________ d) m/COE ________

2. In the diagram to the right, BE AC, BD BF. Find the value of x in each of the following problems.

a) m/ABD = 2x – 15, m/DBE = x x = ________

b) m/ABD = 3x – 12, m/DBE = 2x + 2, m/EBF = 2x + 8 x = ________

>

A B C

,

DF

E

35O

A B

DE

FE

M

RS

A O E

D

B

F

H

C

G

x

y

P

C

D

BA

DB

E

CA

12 4

3

z

A B C

,

DF

E

35O

A B C

DF

E

FE

M

RS

A O E

D

B

F

H

C

G

x

y A B

DE

P

C

BA

DB

E

CA

12 4

3

z

90 + x 180 – y

y + z 90 – x

2x – 15 + x = 90

3x – 15 = 90

3x = 105

x = 35

3x – 12 + 2x + 2 = 905x = 100

x = 20

> >

2x + 2 + 2x + 8 = 904x = 80x = 20

35

20

or


Name

Class Date Score

Quiz Form A

1. The measure of one of two adjacent angles is 10 more than twice the measure of the other. If the sum of their measures is 112, find the measure of each angle. (Note: This is a problem-solving situation, so you may want to use the analysis questions you developed in the Algebra course.)

2. In the diagram, m/1 = 2(m/2), m/2 + m/3 + m/4 = 150,m/1 = m/4, and m/3 = 30. Find m/1, m/2, and m/4.

3. In the diagram, PE Bisects /DPC, m/APC = 72, and /BPD = 70. Find m/APE.

1

2 34

A B

DC

3x136

m

x + 2x + 10 = 112

3x = 102

x = 34

2x + 10 = 78

m/2 + m/3 + m/4 = 150

m/2 + 30 + m/4 = 150

m/2 + m/4 = 120

m/1 = 2(m/2) = m/4

m/2 + 2(m/2) = 120

3m/2 = 120

m/2 = 40

so,

m/1 = 2(m/2) = 80

m/1 = m/4 = 80

72 + x + x + 70 = 180

2x + 142 = 180

2x = 38

x = 19

m/APE = m/APC + m/CPE

m/APE = 72 + 19 = 91

1

2 34

A P B

C E D

DC

P

B

Unit III - Fundamental TheoremsPart E - Theorems About Angles - Part 1 (One Angle)Lesson 1 - Theorem 7: “If, in a half-plane, there is ray in the edge of thehalf-plane then there is exactly one other ray through the endpoint of the givenray, such that the angle formed by the two rays has a given measure.”Lesson 2 - Theorem 8: “If, in a half-plane you have an angle, thenthat angle has exactly one bisector.”

measure of one-angle _________

measure of other angle _________

m/1 = _________

m/2 = _________

m/4 = _________

m/APE _________91O

80O

40O

80O

34O

78O


Unit III, Part E, Lessons 1&2, Quiz Form A—Continued—

Name

4. Given: QT bisects /SQU and

QT PR as shown

Prove: /PQS > /RQU

1. QT bisects /SQU 1. Given

2. /SQT > /UQT 2. Definition of Angle Bisector

3. m/SQT = m/UQT 3. Definition of Congruent Angles

4. QT PR 4. Given

5. /PQT is a right angle 5. Definition of Perpendicular Lines

6. /RQT is a right angle 6. Definition of Perpendicular Lines

7. m/PQT = 90 7. Definition of Right Angle

8. m/RQT = 90 8. Definition of Right Angle

9. m/PQT = m/RQT 9. Substitution (from statements 7 and 8)

10. m/PQS + m/SQT = m/PQT 10. Postulate 7 (Protractor) – Angle-Addition Assumption

11. m/RQU + m/UQT = m/RQT 11. Postulate 7 (Protractor) – Angle-Addition Assumption

12. m/PQS + m/SQT = m/RQU + m/UQT 12. Substitution (from statements 9, 10, 11)

13. m/PQS + m/SQT – m/SQT = m/RQU + 13. Addition for Equalitym/UQT – m/UQT

14. m/PQS + 0 = m/RQU + 0 14. Additive Inverse

15. m/PQS = m/RQU 15. Additive Identity

16. /PQS > /RQU 16. Definition of Congruent Angles

>

>

A P B

C E D

P Q R

S T U

E

D

P

B

CA

PQ

X

R

40110

3

X Y V

R

P T

A B C

D E F

OO

m

T

STATEMENT REASON


Name

Class Date Score

Quiz Form B

Unit III - Fundamental TheoremsPart E - Theorems About Angles - Part 1 (One Angle)Lesson 1 - Theorem 7: “If, in a half-plane, there is ray in the edge of thehalf-plane then there is exactly one other ray through the endpoint of the givenray, such that the angle formed by the two rays has a given measure.”Lesson 2 - Theorem 8: “If, in a half-plane you have an angle, thenthat angle has exactly one bisector.”

1. Draw and label a diagram according to the following instructions:a) Mark points P and A on a line m. Let R and T be the half planes with edge line m.

b) Use a protractor to find a point B in plane R such that m/APB = 110.

c) Choose a point C in half-plane T on PB.

d) measure /APC with a protractor. What is its measure? _____________

e) How would you describe the relationship between /APB and /APC as a pair of angles? ____________

f) How would you describe the relationship between PB and PC as a pair of rays? ____________________

g) What is the sum of m/APB and m/APC? ______________________

2. A point X is in the interior of /PQR. If m/PQX = 40, and m/PQR = 110, what is m/XQR? (Hint: draw andlabel a diagram.)

m/XQR = _________

1

2 34

A P B

C E D

P Q R

S T U

A B E

DC

P

B

CA

PQ

X

R

40110

3x xX Y V

R

P T

A B C

D E F

OO

136

m

T

70O

70O

1

2 34

A P B

C E D

P Q R

S T U

B E

DC

P

B

CA

PQ

X

R

40110

3

P T D E F

OO

136

m

T

Supplementary

Opposite

180O

R


Name

Class Date Score

Quiz Form A

Unit III - Fundamental TheoremsPart F - Theorems About Angles - Part 2 (Two Angles)Lesson 1 - Theorem 9: “If two adjacent acute angles have their exteriorsides in perpendicular lines, then the two angles are complementary.”Lesson 2 - Theorem 10: “If the exterior sides of two adjacent anglesare opposite rays, then the angles are supplementary.”

For each of the following statements 1 through 10, write either true or false.

1. Two angles may be both adjacent and congruent. _______________

2. Two angles may be both complementary and supplementary. _______________

3. If two angles are acute, they cannot be supplementary. _______________

4. If two lines intersect, then four pairs of supplementary and adjacent angles are formed. _______________

5. If /AOB and /BOC are supplementary and adjacent, then OAand OC cannot be a pair of opposite rays. _______________

6. If /AOB and /BOC are adjacent, then B lies inside /AOC. _______________

7. If two angles formed by two lines are adjacent, then they are supplementary. _______________

8. If B lies inside /AOC, then /AOB and /BOC are adjacent. _______________

9. If the measure of an angle is 120, the measure of the complement is 60. _______________

10. If /AOB and /BOC are adjacent, then m/AOB + m/BOC = m/AOC. _______________

True

False

True

True

False

True

True

True

False

True


Unit III, Part F, Lessons 1&2, Quiz Form A—Continued—

Name

11. Given: /DOC is a complement of /BOC and/AOB is a complement of /BOC as shown

Prove: m/AOB = m/DOC

1. /DOC is a complement of /BOC 1. Given

2. /AOB is a complement of /BOC 2. Given

3. m/DOC + m/BOC = 90 3. Definition of Complementary Angles

4. m/AOB + m/BOC = 90 4. Definition of Complementary Angles

5. m/DOC + m/BOC = m/AOB + m/BOC 5. Substitution of Equals

6. m/BOC = m/BOC 6. Reflexive Property of Equality

7. m/DOC + m/BOC – m/BOC = m/AOB + 7. Subtraction Property of Equalitym/BOC – m/BOC

8. m/DOC + m/BOC + —m/BOC = m/AOB + 8. Definition of Subtraction (a–b means a + —b)m/BOC + —m/BOC

9. m/DOC + 0 = m/AOB + 0 9. Additive Inverse Property

10. m/DOC = m/AOB 10. Identity Property of Addition

11. m/AOB = m/DOC 11. Symmetry Property of Equality

A O

BC D

D

A

C

U

V

T

W

X

Y

R

T

M 31 2

4Q

21

,

,1

2

B

A

DC

O

B

CA D

X

STATEMENT REASON

NameUnit III, Part F, Lessons 1&2, Quiz Form A—Continued—


12. Given: /ADC > /BEC as shown

Prove: /BDC > /AEC

1. /ADC and /CDB are adjacent angles 1. Definition of Adjacent Angle/BEC and /CEA are adjacent angles

2. DA and DB are opposite rays 2. Definition of Opposite RaysEB and EA are opposite rays

3. /ADC and /CDB are supplementary 3. Theorem 10 - If the exterior sides of two adjacent angles are/BEC and /CEA are supplementary opposite rays, then the angles are supplementary.

4. m/ADC + m/CDB = 180 4. Definition of Supplementary Anglesm/BEC + m/CEA = 180

5. m/ADC + m/CDB = m/BEC + m/CEA 5. Substitution of Equals (4 into 4)

6. /ADC > /BEC 6. Given

7. m/ADC = m/BEC 7. Definition of Congruent Angles

8. m/ADC + m/CDB – m/ADC = m/BEC + 8. Addition Property of Equalitym/CEA – m/BEC

9. m/ADC + m/CDB + —m/ADC = m/BEC + 9. Definition of Subtraction (a – b means a + b)m/CEA + —m/BEC

10. m/ADC + —m/ADC + m/CDB = m/BEC + 10. Commutative Property of Addition—m/BEC + m/CEA

11. 0 + m/CDB = 0 + m/CEA 11. Additive Inverse Property

12. m/CDB = m/CEA 12. Identity Property of Addition

13. /BDC > /AEC 13. Definition of Congruent Angles

A O

BC D

D E

BA

C

U

V

T

W

X

Y

R

T

NM 31 2

4Q

X

Z

YW

21

,

,1

2

B

A

F

E

DC

OXD

A

B

CA D

Q

A B

1 2

X Y

A

E

B

STATEMENT REASON


Name

Class Date Score

Quiz Form B

Unit III - Fundamental TheoremsPart F - Theorems About Angles - Part 2 (Two Angles)Lesson 1 - Theorem 9: “If two adjacent acute angles have their exteriorsides in perpendicular lines, then the two angles are complementary.”Lesson 2 - Theorem 10: “If the exterior sides of two adjacent anglesare opposite rays, then the angles are supplementary.”

1. Find the measures of /A and /B if m/B = 9m/A and if they are:

a) complementary: m/A = _________ b) supplementary: m/A = _________

m/B = _________ m/B = _________

2. Find the measures of /A and /B if m/B = 28 + m/A and if they are:

a) complementary: m/A = _________ b) supplementary: m/A = _________

m/B = _________ m/B = _________

3. May /AOB and /BOC be adjacent if m/AOB + m/BOC = x, where

a) x = 90 ? _________ b) x = 180 ? _________ c) x < 180 ? _________

d) x > 180 ? _________ e) x < 360 ? _________ f) x = 360 ? _________

m/A + m/B = 90

m/B = 9m/A

m/A + 9m/A = 90

10m/A = 90

m/A = 9

m/B = 81

m/A + m/B = 180

m/B = 9m/A

m/A + 9m/A = 180

10m/A = 180

m/A = 18

m/B = 162

m/A + m/B = 90

m/B = 28 + m/A

m/A + 28 + m/A = 90

2m/A + 28 = 90

2m/A = 62

m/A = 31 m/B = 59

m/A + m/B = 180

m/B = 28 + m/A

m/A + 28 + m/A = 180

2m/A + 28 = 180

2m/A = 152

m/A = 76 m/B = 104

yes yes yes

yes yes yes

9O

81O

18O

162O

31O

59O

76O

104O


Name

Class Date Score

Quiz Form A

Unit III - Fundamental TheoremsPart F - Theorems About Angles - Part 2 (Two Angles)Lesson 3 - Theorem 11: “If you have right angles, then those right anglesare congruent.”Lesson 4 - Theorem 12: “If you have straight angles, then thosestraight angles are congruent.”

1. Name each of the following using the figure at the right.

a) Two pairs of opposite rays. _______________

b) Two right angles. _______________

c) Two straight angles. _______________

d) Three acute angles. _______________

e) Three obtuse angles. _______________

f) Two points in the exterior of /VYT _______________

g) The sides of /XYV _______________

h) The vertex of all angles. _______________

i) A point in the interior of /TYV _______________

j) An angle which is congruent to /WYV _______________

k) An angle which is congruent to /TYW _______________

YU and YX; YT and YW

/VYW; /VYT

/TYW; /XYU

/TYU; /UYV; /XYW

/XYT; /UYW; /XYV

point X; point W

YX; YV

point Y

point U

/VYT

/XYU

A O

BC D

D

A

U

V

T

W

X

Y

21

,

,1

2

B

CA D

NameUnit III, Part F, Lessons 3&4, Quiz Form A—Continued—


5. Given: /XYD is a right angle, /ABW is a right angle,/XYZ is a straight angle, and /ABC is a straight angle as shown

Prove: /DYZ and /WBC are supplementary

1. /XYD is a right angle 1. Given2. /ABW is a right angle 2. Given

3. m/XYD = 90 3. Definition of a Right Angle4. m/ABW = 90 4. Definition of a Right Angle

5. /XYZ is a straight angle 5. Given6. /ABC is a straight angle 6. Given

7. m/XYZ = 180 7. Definition of a Straight Angle8. m/ABC = 180 8. Definition of a Straight Angle

9. m/XYZ = m/XYD + m/DYZ 9. Postulate 7 (Protractor) – Angle-Addition Assumption10. m/ABC = m/ABW + m/WBC 10. Postulate 7 (Protractor) – Angle-Addition Assumption

11. 180 = 90 + m/DYZ 11. Substitution of Equals12. 180 = 90 + m/WBC 12. Substitution of Equals

13. 180 – 90 = 90 + m/DYZ – 90 13. Subtraction Property for Equality14. 180 – 90 = 90 + m/WBC – 90 14. Subtraction Property for Equality

15. 90 = 90 + m/DYZ – 90 15. Substitution of Equals16. 90 = 90 + m/WBC – 90 16. Substitution of Equals

17. 90 = 90 + m/DYZ + –90 17. Definition of Subtraction18. 90 = 90 + m/WBC + –90 18. Definition of Subtraction

19. 90 = 90 + –90 + m/DYZ 19. Commutative Property of Addition20. 90 = 90 + –90 + m/WBC 20. Commutative Property of Additio

21. 90 = 0 + m/DYZ 21. Additive Inverse Property22. 90 = 0 + m/WBC 22. Additive Inverse Property

23. 90 = m/DYZ 23. Identity Property of Addition24. 90 = m/WBC 24. Identity Property of Addition

25. 90 + 90 = m/DYZ + m/WBC 25. Addition Property for Equality

26. 180 = m/DYZ + m/WBC 26. Substitution of Equals (90 + 90 = 180)

27. /DYZ and /WBC are supplementary 27. Definition of Supplementary Angles

STATEMENT REASON

A O

BC D

D E

BA

C

U

V

T

W

X

Y

R

T

NM 31 2

4Q

X

Z

Y

A

C

B DW

21

,

,1

2

B

A

F

E

DC

OX BD

A

B

CA D

Q

A B

1 2

X Y

DA

E

C

B

A

CB

E

A– 4– 5 – 3 – 2 – 1 0 1

– 3– 4 – 2 – 1 0 1 2


Name

Class Date Score

Quiz Form B

Unit III - Fundamental TheoremsPart F - Theorems About Angles - Part 2 (Two Angles)Lesson 3 - Theorem 11: “If you have right angles, then those right anglesare congruent.”Lesson 4 - Theorem 12: “If you have straight angles, then thosestraight angles are congruent.”

1. Name each of the following using the figure at the right.

a) Three right angles. _______________

b) Two pairs of opposite rays. _______________

c) Two rays that are not opposite rays. _______________

d) Two straight angles. _______________

e) All acute angles. _______________

f) Four obtuse angles. _______________

g) Two points in the interior of /BOF _______________

h) The sides of /AOC. _______________

i) An angle which is congruent to /AOE _______________

j) An angle which is congruent to /COF _______________

k) Two points in the exterior of /COE. _______________

/COD; /DOF; /BOE; /AOE

OC and OF; OB and OA

OD; OE

/COF; /BOA

/COB; /BOD; /DOE; /EOF; /FOA

/COA; /COE; /BOF; /DOA

point D; point E

OA; OC

/EOB, /DOC, /DOF

/BOA

point F; point A

A O

BC D

D E

BA

U

V

T

W

X

Y

R

T

NM 31 2

4QW

21

,

,1

2

B

A

F

E

DC

O

B

CA D

Q

A

1 2

X

A

NameUnit III, Part F, Lessons 3&4, Quiz Form B—Continued—


2. Solve the inequality 2y + 8 < – 6y, and graph the solution on a number line. Then answer the following:

a) Does the graph of the solution set of the inequality represent a ray?__________ Explain. __________

__________________________________________________________________________________

b) Write the inequality whose solution set is the complement of the solution set for 2y + 8 < –6y.

__________________________________________________________________________________

c) The union of the two solutions to the inequalities in part a) and part b) is a ________________ angle.straight

No the graph

does not have a definite endpoint

2y + 8 < –6y

2y + –2y + 8 < –6y + –2y

0 + 8 < –8y

8 < –8y

–1 > y or y < –1

1

–88 >

1

–8⋅

⋅ −8y

2y + 8 < –6y or 2y + 8 $ –6y

A

F

E

D

OX BD

Q

A B

1 2

X Y

CA– 4– 5 – 3 – 2 – 1 0 1

– 3– 4 – 2 – 1 0 1 2


Name

Class Date Score

Quiz Form A

Unit III - Fundamental TheoremsPart G - Theorems About Angles - Part 3 (More Than 2 Angles)Lesson 1 - Theorem 13: “If two angles are complementary to the sameangle or congruent angles, then they are congruent to each other.”Lesson 2 - Theorem 14: “If two angles are supplementary to the sameangle or congruent angles, then they are congruent to each other.”Lesson 3 - Theorem 15: “If two lines intersect, then the vertical anglesformed are congruent.”

In the diagram, at the right, /AFB is a right angle. Name the figures described in exercises 1 through 6 below.

1. Another right angle _______________

2. Two complementary angles. _______________

3. Two congruent supplementary angles. _______________

4. Two non-congruent supplementary angles. _______________

5. Two acute vertical angles. _______________

6. Two obtuse vertical angles. _______________

In the diagram, at the right, OT bisects /SOU, m/UOV = 30 and m/YOW = 126. Find the measure of each angle.

7. m/VOW _______________

8. m/ZOY _______________

9. m/TOU _______________

10. m/ZOW _______________

11. m/UOS _______________

12. m/TOZ _______________

/AFE and /EFD

/BFA and /DFA/CFA and /EFA or/BFC and /DFC or

/BFE and /EFD

/BFC and /DFE

/CFD and /BFE

24O

30O

63O

156O

126O

87O

DFB

C

E

A

S

x 2x+12P

B

C

B D

C

A

1

2

3 4

/AFD

DFB

C

E

A

WS O

XY

Z

TU

V

x 2x+12P

F

E

B

C

D

A


Name

Class Date Score

Quiz Form B

Unit III - Fundamental TheoremsPart G - Theorems About Angles - Part 3 (More Than 2 Angles)Lesson 1 - Theorem 13: “If two angles are complementary to the sameangle or congruent angles, then they are congruent to each other.”Lesson 2 - Theorem 14: “If two angles are supplementary to the sameangle or congruent angles, then they are congruent to each other.”Lesson 3 - Theorem 15: “If two lines intersect, then the vertical anglesformed are congruent.”

In the diagram, at the right, PA CF and PD BE. Refer to this diagram for exercises 1 through 6.

1. Find two supplementary angles for /FPE. ___________________

2. Find two complementary angles for /BPC. __________________

3. /APB > /CPD. Why? _________________________________

_________________________________________________________

_________________________________________________________

4. /BPF > /CPE. Why? _________________________________

_________________________________________________________

_________________________________________________________

5. /BPC > /FPE. Why? _________________________________

_________________________________________________________

_________________________________________________________

6. If m/BPC = 35, then

a) m/CPD = ______________

b) m/FPE = ______________

c) m/APB = ______________

7. If /A is complementary to /B, and if the measure of the supplement of /B is 122, Find m/A.

m/A = _________

DFB

C

E

A

WS O

XY

Z

TU

V

x 2x+12P

F

E

B

C

D

A

B D

C

A

1

2

3 4 A BC

/FPB and /CPE

/BPA and /DPC

Theorem 13 - If two angles are

Theorem 14 - If two angles are supplementary

to the same angle, then they are congruent to each other. or Theorem 15 - If two lines

intersect, then the vertical angles formed are congruent.

complementary to the same angle, then they are congruent to each other.

Theorem 15 - If two lines intersect, then the

vertical angles formed are congruent.

55O

35O

55O

The supplement of /B is 122. Therefore, m/B = 58O. /A is complementary to /B.

Therefore, m/A = 32O

32O

> >


Name

Class Date Score

Quiz Form A

Unit III - Fundamental TheoremsPart H - Theorems About Parallel LinesLesson 1 - Postulate 11: “If teo parallel lines are cut by a transversal,then corresponding angles are congruent.”Lesson 2 - Theorem 16: “If two parallel lines are cut by a transversal,then alternate interior angles are congruent.”Lesson 3 - Theorem 17: “If two parallel lines are cut by a transversal,then interior angles on the same side of the transversal are supplementary.”

1. Complete the following conditional by writing three different true conclusions as given by Postulate 11, Theorem 16, and Theorem 17.

If two parallel lines are cut by a transversal, then

a) ____________________________________________________________

b) ____________________________________________________________

c) ____________________________________________________________

Use the diagram at the right, and only the numbered angles, for Exercises 2, 3, 4, and 5.

2. Using t1 and t2 as transversals, name all the pairs of corresponding angles.

3. Using ,1 and ,2 as transversals, name all the pairs of corresponding angles.

4. Using t1 and t2 as transversals, name all the pairs of interiorangles on the same side of the transversal.

5. Using ,1 and ,2 as transversals, name all the pairs of vertical angles.

Corresponding Angles are congruent

Alternate Interior Angles are congruent

Interior angles on the same side of the transversal are supplementary

A

1 23 4 9 10

5 6 7 8

t1 t2

,1

,2

PN

M

S

RQ

70

(3x – 14)

,1

,291

(5x – 24)

t

t

0

0

0

0

/1 and /5

/2 and /6

/3 and /9

/4 and /10

/5 and /7

/6 and /8

/3 and /5

/4 and /6

/9 and /7

/10 and /8

/1 and /4

/2 and /3


NameUnit III, Part H, Lessons 1, 2 & 3, Quiz Form A—Continued—


15. Given: QR || PT as shown

Prove: /4 is supplementary to /3

,2

(3x – 14)(2x – 10)

,1

,2

,1 ,2

(2x – 8),1

,2

(15x – 19) (7x – 3)

T

R

M

NP

Q

32

1 4

c

d

p

q

4 13 2

12 911 10

8 57 6

16 1315 14

t r

B

AE

C

D 1 23

BA

E D C

E

B1 2

3 4B

D

A

C 5 67 8

t r

D

A

B

E

C

F

12

3

,1

,2

(12x – 9) 135

C3

CA B

F

E

D

D GB

FE

A C

m

t

,

1 2

5 6

3 4

7 8

W

E F

G H

A

B

CE

D

A B

D C

1 ,1 A

I

H E

A

B1

t

t

0

0

0 0

0

0 0

1. QR || PT 1. Given

2. /1 and /2 are supplementary 2. Theorem 17 - If two parallel lines are cut by a transversal, then interior angles on the same side of the transversal are supplementary

3. m/1 + m/2 = 180 3. Definition of Supplementary Angles

4. /1 > /3 4. Postulate 11 - If two parallel lines are cut by a transversal, then corresponding angles are congruent.

5. /2 > /4 5. Theorem 16 - If two parallel lines are cut by a transversal, then alternate interior angles are congruent

6. m/1 = m/3 6. Definition of Congruent Angles


8. m/3 + m/4 = 180 8. Substitution of Equality (6 and 7 into 3)

9. /4 is supplementary to /3 9. Definition of Supplementary Angles

STATEMENT REASON


Name

Class Date Score

Quiz Form B

Unit III - Fundamental TheoremsPart H - Theorems About Parallel LinesLesson 1 - Postulate 11: “If teo parallel lines are cut by a transversal,then corresponding angles are congruent.”Lesson 2 - Theorem 16: “If two parallel lines are cut by a transversal,then alternate interior angles are congruent.”Lesson 3 - Theorem 17: “If two parallel lines are cut by a transversal,then interior angles on the same side of the transversal are supplementary.”

Use the figure at the right to name all of the pairs of angles asked forin Exercises 1 through 4.

For these And this Name all pairs of Name all pairs of Name all pairs of Interior Angles

Lines Transversal Alternate Interior Angles Corresponding Angles of the same side of the transversal

1. c, d e _____________________ ___________________ ___________________

_____________________ ___________________ ___________________

2. c, d f _____________________ ___________________ ___________________

_____________________ ___________________ ___________________

3. e, f c _____________________ ___________________ ___________________

_____________________ ___________________ ___________________

4. e, f d _____________________ ___________________ ___________________

_____________________ ___________________ ___________________

PN

M

S

RQ

70

(3x – 14)

,1

,291

(5x – 24)

1 25 6 7 8

3 4

9 1013 14

11 1215 16

e f

c

d

3 1

2 E

B

D

A

C

A

C7

E

D C

BA

13

2

A

F

S

W

T

R

Z

YX

QP

NM E

t

t

0

0

0

/5 and /10; /6 and /9

/7 and /12; /8 and /11

/2 and /7; /3 and /6

/10 and /15; /11 and /14

/1 and /9; /5 and /13

/3 and /11; /7 and /15

/1 and /3; /2 and /4

/9 and /11; /10 and /12

/2 and /10; /6 and /14

/4 and /12; /8 and /16

/5 and /7; /6 and /8

/13 and /15; /14 and /16

/5 and /9; /6 and /10

/7 and /11; /8 and /12

/2 and /3; /6 and /7

/10 and /11; /14 and /15

NameUnit III, Part H, Lessons 1,2&3, Quiz Form B—Continued—


In the diagram at the right, p || q and t || r. Use this diagramto find m/12 in Exercises 5 through 12.

5. m/2 = 75O m/12 = _________ 6. m/3 = 110O m/12 = _________

7. m/16 = 80O m/12 = _________ 8. m/14 = 72O m/12 = _________

9. m/6 = 68O m/12 = _________ 10. m/5 = 104O m/12 = _________

11. m/12 + m/16 = 132O m/12 = ________ 12. m/8 + m/10 = 162O m/12 = _________

,291

(5x – 24)

1 25 6 7 8

3 4

9 1013 14

11 1215 16

e f

c

d

p

q

4 13 2

12 911 10

8 57 6

16 1315 14

t r

B

A

3 1

2 E

B

D

A

C

1 23 4

B

D

A

C 5 67 8

t r

B

E

1

E

D C

BA

13

2

CA B

F

E

D

D

E

A

S

W

T

R

Z

YX

QP

NM E F

G H

A

B

ED

B

CE

G F

A

D

1

2

3

,1

,2

,3

t

F

G B

J

EA

G

C

t

0

0

m/2 = m/12 m/3 = m/1

m/12 + m/1 = 180

m/12 + 110 = 180

m/12 = 70

m/16 = m/12 m/14 = m/16

m/16 = m/12

m/14 = m/12

m/6 = m/16

m/16 = m/12

m/6 = m/12

m/5 + m/16 = 180

m/16 = m/12

m/5 + m/12 = 180

104 + m/12 = 180

m/12 = 76

m/12 = m/16

m/12 = 66

m/8 = m/16

m/16 = m/10

m/8 = m/10

m/10 = 81

m/10 = m/12

m/12 = 81

70O

80O 72O

68O 76O

66O 81O

75O


Name

Class Date Score

Quiz Form A

Unit III - Fundamental TheoremsPart H - Theorems About Parallel LinesLesson 4 - Theorem 18: “If a given line is perpendicular to one of twoparallel lines, then it is perpendicular to the other.”Lesson 5 - Theorem 19: “If two lines are cut by a transversal so thatcorresponding angles are congruent, then the two lines are parallel.”Lesson 6 - Theorem 20: “If two lines are cut by a transversal so thatalternate interior angles are congruent, then the two lines are parallel.”Lesson 7 - Theorem 21: “If two lines are cut by a transversal so thatinterior angles on the same side of the transversal are supplementary, thenthe two lines are parallel.”

Refer to the given diagram at the right and the given information inExercises 1 through 9. State the reason to justify AB || CD.

1. m/3 = 50, m/6 = 50

2. m/1 = 130, m/8 = 130

3. m/4 = 100, m/6 = 80

4. m/1 = 120, m/7 = 60

5. m/7 = 55, m/3 = 55

6. m/2 = 65, m/7 = 65

7. m/6 = 57, m/2 = 57

8. m/4 = 110, m/5 = 110

9. Why is line r AB?

GF

C

D

BA

E

1 23 4 9 10

5 6 7 8

t1 t2

,1

,2

B

A

E D1 2

3 4

PN

M

S

RQ

,1

,2

70

(3x – 14)(2x – 10)

,1 ,2

,1

,291

(5x – 24)

,1 ,2

(2x – 8),1

,2

(15x – 19) (7x – 3)

P

Q

1 25 6 7 8

3 4

9 1013 14

11 1215 16

e f

c

d

p

q

4 13 2

12 911 10

8 57 6

16 1315 14

t r

B

AE

C

D

A

3 1

2 E

B

D

A

C

1 23 4

B

D

A

C 5 67 8

t r

D

A

B

E

C

F

12

3

(12x –

E

D C

BA

13

2

CA B

F

E

D

D GB

FE

A C

1 2

5 6

S

W

T

R

Z

YX

QP

NM E F

G H

A

B

CE

DD

B

CE

G F

A

D

1

2

3

,1

,2

,3

t

EF

G B

J

D

C

A

I

H E

A

C43

21

E

F

A

B

G

HD

C

t

t

t

t

t

0

0

0

0

0

0

0 0

Theorem 20 – If two lines are cut by a transversal so that alternate interior angles are congruent,

then the two lines are parallel.

Corollary 20a – If two lines are cut by a transversal so that alternate exterior angles

are congruent, then the two lines are parallel.

Theorem 21 – If two lines are cut by a transversal so that interior angles on the same side of the

transversal are supplementary, then the two lines are parallel.

Corollary 21a – If two lines are cut by a transversal so that exterior angles on the same

side of the transversal are supplementary, then the two lines are parallel.

Theorem 19 – If two lines are cut by a transversal so that corresponding angles are congruent,


Corollary 20a – If two lines are cut by a transversal so that alternate exterior angles

are congruent, then the two lines are parallel.

Theorem 19 – If two lines are cut by a transversal so that corresponding angles are congruent,


Theorem 20 – If two lines are cut by a transversal so that alternate interior angles are congruent,

then the two lines are parallel

Theorem 18 – If a given line is perpendicular to one of two parallel lines, then it is perpendicular

to the other.>

NameUnit III, Part H, Lessons 4,5,6&7, Quiz Form A—Continued—


13. Given: DG || AC and/ACE is a right angle as shown

Prove: /ABC and /FBG are complementary angle

2 EDC

DC 5 67 8

t r

B

E

C

F

12

3

E

D C

BA

13

2

CA B

F

E

D

D GB

FE

A C

,

1 2

5 6

3 4

7 8

S

W

T

R

Z

YX

QP

NM E F

G H

A

B

CE

D

A

D

B

CE

G F

A

D

1

2

3

,1

,2

,3

t

EF

G B

J

D

C

A

I

H E

A

B

CF

D

43

21

E

F

A

B

G

HD

C

1. /ACE is a right angle 1. Given

2. CE AC 2. Definition of Perpendicular Lines.

3. DG || AC 3. Given

4. CE DG 4. Theorem 18 - If a given line is perpendicular to one of two parallel lines, then it is perpendicular to the other.

5. /DBC is a right angle 5. Definition of Perpendicular Lines

6. m/DBC = 90 6. Definition of Right Angle.

7. m/DBA + m/ABC = m/DBC 7. Postulate 7 (Protractor) - Angle-Addition Assumption

8. /DBA > /FBG 8. Theorem 15 - If two lines intersect, then the vertical angles formed are congruent

9. m/DBA = m/FBG 9. Definition of Congruent Angles

10. m/FBG + m/ABC = m/DBC 10. Substitution of Equality (9 into 7)

11. m/FBC + m/ABC = 90 11. Substitution of Equality (6 into 10)

12. /ABC and /FBC are complementary angles 12. Definition of Complementary Angles

STATEMENT REASON

>

>


Name

Class Date Score

Quiz Form B

Unit III - Fundamental TheoremsPart H - Theorems About Parallel LinesLesson 4 - Theorem 18: “If a given line is perpendicular to one of twoparallel lines, then it is perpendicular to the other.”Lesson 5 - Theorem 19: “If two lines are cut by a transversal so thatcorresponding angles are congruent, then the two lines are parallel.”Lesson 6 - Theorem 20: “If two lines are cut by a transversal so thatalternate interior angles are congruent, then the two lines are parallel.”Lesson 7 - Theorem 21: “If two lines are cut by a transversal so thatinterior angles on the same side of the transversal are supplementary, thenthe two lines are parallel.”

1. To prove two lines in the same plane are parallel, if they are cut by a transversal, you can show that one of

five conditions is true. List the five conditions in a) through e) below. (three theorems; two corollaries)

a)

b)

c)

d)

e)

If a pair of alternate interior angles are congruent.

If a pair of corresponding angles are congruent.

If a pair of alternate exterior angles are congruent.

If a pair of interior angles on the same side of the transversal are supplementary.

If a pair of exterior angles on the same side of the transversal are supplementary.


NameUnit III, Part H, Lessons 4,5,6&7, Quiz Form B—Continued—

11. Given: AB || DF and/BAC > /GFD as shown

Prove: GF || AC

WR YX

QP

GB

CE

G F

A

D

1

2

3

t

E

F

A

B

G

HD

C

1. AB || DF 1. Given

2. /BAC > /FEC 2. Postulate 11 - If two parallel lines are cut by a transversal, then corresponding angles are congruent.

3. m/BAC = m/FEC 3. Definition of Congruent Angles

4. /BAC > /GFD 4. Given

5. m/BAC = m/GFD 5. Definition of Congruent Angles

6. m/FEC = /GFD 6. Substitution of Equals (3 into 5)

7. /FEC > /GFD 7. Definition of Congruent Angles

8. GF || AC 8. Theorem 20 - If two lines are cut by a transversal so that alternate interior angles are congruent, then the two lines are parallel.

STATEMENT REASON


Name

Class Date Score

Quiz Form A

Unit III - Fundamental TheoremsPart H - Theorems About Parallel LinesLesson 8 - Theorem 22: “If two lines are perpendicular to a third line,then the two lines are parallel to each other.”Lesson 9 - Theorem 23: “If two lines are parallel to a third line, thenthe two lines are parallel to each other.”Lesson 10 - Theorem 24: “If two parallel planes are cut by a third plane, then the two lines of intersection are parallel.”

In Exercises 1 through 10, classify each statement as always, sometimes, or never true. (circle your choice)

1. Two planes, each perpendicular to a third plane, are Always Sometimes Never True parallel to each other.

Counter example:

2. A plane that cuts one of two parallel lines cuts the other also. Always Sometimes Never True

illustration:

3. In a plane, a line which intersects one of two parallel lines, Always Sometimes Never Trueintersects the other also.

illustration:

4. A plane that contains two sides of a triangle contains the Always Sometimes Never Truethird side also. (Remember: 3 points will determine a plane)

illustration:

t

,

P

Q

t

,

P

Q

t

,

P

Q

t

,

P

Q

NameUnit III, Part H, Lessons 8,9&10, Quiz Form A—Continued—


12. Given: DF CI andCG CI as shown

Prove: /FEH > /CBJ

WYX

Q

G H

CE

DD

B

CE

F1

2

3

,1

,2

,3

t

EF

G B

J

D

C

A

I

H E

A

B

CF

D

43

21

F H

1. DF CI 1. Given

2. CG CI 2. Given

3. DF || CG 3. Theorem 22 - If two lines are perpendicular to a third line, then the two lines are parallel.

4. /FEH > /CBJ 4. Corollary 16a - If two parallel lines are cut by a transversal, then alternate exterior angles are congruent.

STATEMENT REASON

>>

>>


Name

Class Date Score

Quiz Form B

Unit III - Fundamental TheoremsPart H - Theorems About Parallel LinesLesson 8 - Theorem 22: “If two lines are perpendicular to a third line,then the two lines are parallel to each other.”Lesson 9 - Theorem 23: “If two lines are parallel to a third line, thenthe two lines are parallel to each other.”Lesson 10 - Theorem 24: “If two parallel planes are cut by a third plane, then the two lines of intersection are parallel.”

In Exercises 1 through 10, classify each statement as always, sometimes, or never true. (circle your choice)

1. If two planes are parallel, every line in one of the planes is Always Sometimes Never True parallel to the other plane.

illustration:

2. In space, a line which intersects one of two parallel lines, Always Sometimes Never Trueintersects the other also.

Counter example:

3. A line and a plane are parallel if they do not intersect. Always Sometimes Never True

illustration:

4. If line , is parallel to plane P, and plane Q contains line ,, Always Sometimes Never Trueand plane P intersects plane Q in line t, then line t isparallel to line ,.

illustration:

t

,

P

Q

t

,

P

Q

t

,

P

Q

NameUnit III, Test Form A—Continued—


32. Given: /DEB and /TBE are right angles/1 > /4

Prove: /2 > /3

1. /DEB and /TBE are right angles 1. Given

2. /DEB > /TBE 2. Theorem 11 - If you have right angles, then those right anglesare congruent.

3. m/DEB = m/TBE 3. Definition of Congruent Angles

4. m/DEB = m/1 + m/2 4. Postulate 7 (Protractor) - Angle-Addition Assumption

5. m/TBE = m/3 + m/4 5. Postulate 7 (Protractor) - Angle-Addition Assumption

6. /1 > /4 6. Given


8. m/1 + m/2 – m/1 = m/3 + m/4 – m/4 8. Addition (or Subtraction) Property for Equality

9. m/2 + m/1 – m/1 = m/3 + m/4 – m/4 9. Commutative Property of Addition

10. m/2 + 0 = m/3 + 0 10. Additive Inverse Property

11. m/2 = m/3 11. Identity Property for Addition

12. /2 = /3 12. Definition of Congruent Angles

STATEMENT REASON

Q

t

5 67 8

,

ts

m

1

2

3 45 6

7 89 10

,2

,3

,1

1 23 4

75 6

8 E B

D T

12 3

4

1

2

34

m

n

1 23 4

m5 67 8

,

t

,2

,3

,1

1 23 4

75 6

8

A T

R F

A B

C D

E F

1

2

E

B

A

F

E

A

D

B

H

3

2BD

EC

G F

1

p

q

2

431

RM S

,1 ,2

t A B C

D C D

P QN

A

M

B

P Q

T

Q,

m

(II-C-1)(III-F-3)

NameUnit III, Test Form B—Continued—


2. Given: collinear points, P, Q, R, and SPQ > SR

Prove: PR > SQ

1. Points P, Q, R and S are collinear 1. Given

2. PQ > SR 2. Given

3. PQ = SR 3. Definition of Congruent Line Segments

4. QR = RQ 4. Reflexive Property of Equality

5. PQ + QR = SR + RC 5. Addition Property for Equality

6. PQ + QR = PR 6. Postulate 6 (Ruler) - Segment-Addition Assumption

7. SR + RC = SQ 7. Postulate 6 (Ruler) - Segment-Addition Assumption

8. PR = SQ 8. Substitution of Equality (statements 5, 6 and 7)

9. PR > SQ 9. Definition of Congruent Line Segments

STATEMENT REASON

,2

,3

,1

1 23 4

75 6

8

A T

R F

A B

C D

E F

1

2

F

E

G

D

B

H

qM S

B C

D C D

P QN

A

M

B

P Q R S

T

Q,

m

(II-C-1,7)

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