1. Measuring Segments and Angles To measure segments we use a ruler. 123456 ●● In this example, the length or measure of line segment AB is 2 inches We.

1

Measuring Segments and AnglesMeasuring Segments and Angles

To measure segments we use a ruler.

11 22 33 44 55 66

●● ●●

In this example, the length or measure of line segment AB is 2 inches

We write this as mAB We cannot write AB = 2

INCHESINCHES

AA BB

2

2.12.1

Since the number of points between A and B are infinite and NOT limited to 2.

Measuring Segments and AnglesMeasuring Segments and Angles

To measure angles we use a protractor.

●

●

●●

●

●●●AAPPFF

ZZ

BBWW

SS

For example, the measure of angle ZPA is 50 ° written as m< ZPA = 50 ° 3

2.12.1

Classifying AnglesClassifying Angles

Acute Angle ( < 90 ) Right Angle ( = 90 ) Obtuse Angle ( > 90 )

A is between 0 and 90( 0 < a < 90 ) A = 90 A is between 90 and 180

( 90 < a < 180 )

a °a ° a °

4

2.12.1

Betweenness of Points and RaysBetweenness of Points and Rays

Definition of Betweeness:Point C is between points A and B if both the following conditions are met:1.Points A, C, and B are three different collinear points2.AB = AC + CB

●● ●● ●●AA BBCC

22 33

Example: Point C is between A and B. If AC = 2 and CB = 3, find AB

60 °

20 °40 °

AA

BB

PP

OO

●●

●●

●●

●●

If m < AOP = 40 and m < POB = 20 then m < AOB = 60. This logical relationship is called:

Angle Addition PostulateAngle Addition Postulate

If ray OP is in the interior of <AOB, then m < AOP = m < AOP + m < POB

5

2.12.1

INITIAL POSTULATESINITIAL POSTULATES

In building a geometric system, not everything can be proved since there must be some basic assumptions, called “postulates” or axioms, that are needed as a beginning.

POSTULATE 1.1 POSTULATE 1.1 Two points determine a lineTwo points determine a line

● ●A B

POSTULATE 1.2 POSTULATE 1.2 Three non-collinear points determine a planeThree non-collinear points determine a plane

A●

B

●

●

C6

2.12.1

BG lies in the interior of < ABC.If m < ABG = 30 and m < GBC = 20, then find m < ABC 20 °

30 °

AA

CC

GG

BB

●●

●●

●●

●●

7

Example 1 of Angle Addition PostulateExample 1 of Angle Addition Postulate

??

2.22.2

KM lies in the interior of < JKL.If m < JKL = 50 and m < MKL = 20, then find m < ABC

20 °

50 °

JJ

LL

MM

KK

●●

●●

●●

●●

The Angle Addition Postulate may also be expressed as m < JKM = m < JKL - m < MKL

8

Example 2 of Angle Addition PostulateExample 2 of Angle Addition Postulate

??

2.22.2

130 °

90 °

GG

CC

AA

BB

●●

●●

●●

●●

9

Example 3 of Angle Addition PostulateExample 3 of Angle Addition Postulate

??

140 °

BG is not in the interior of < ABC, thus violating the assumption (or hypothesis ) of the Angle Addition Postulate.

REASON:

Interior of < ABC

2.22.2

10

CongruenceCongruence

Figures that have the same shape and size are said to be CONGRUENT.CONGRUENT.

REMEMBER: Figures are congruent only if they agree in all their dimensions.REMEMBER: Figures are congruent only if they agree in all their dimensions.

4

4 4

4

4

4

4 4Same size but not the same shape

2

Same shape but not the same size

2

2

2

If line segments have the same length, they are congruent. Notation: AB RS

If 2 angles have the same measure , they are congruent. m < ABC m < DEF

2.32.3

11

Midpoint and BisectorMidpoint and Bisector

●●●● ●●BBAA MM

●● ●●●●AA BB

MM

XX

YY ●●

●●

DEFINITION OF MIDPOINT Point M is the midpoint of AB if1.M is between A and B2.AM = MB

DEFINITION OF A SEGMENT BISECTOR A bisector of a line segment AB is any line, ray, or segment that passes through the midpoint of AB. Thus, a segment bisector divides a segment into 2 congruent segments.

33 33

2.32.3

12

The midpoint of the line segment joining

A (x1, y1) and B (x2, y2) is as follows:

•

•

•A ( ─2, 5)

C ( 1, 3)

B ( 4, 1)

M = (x1 + x2

,y1 + y2

)2 2Each coordinate of M is the mean of the corresponding coordinates of A and B.

M = (─2 + 4

,5 + 1

)2 2

M = (2

,6

)2 2

The Midpoint FormulaThe Midpoint FormulaThe Midpoint FormulaThe Midpoint Formula 1.41.4

13

The Distance FormulaThe Distance FormulaLet A = ( x1 , y1) and B = ( x2 , y2 ) be points in a coordinate plane. The distance between A and B is

AB = ( x2 ─ x1 ) 2 + ( y2 ─ y1 ) 2

Example 1 Using the Distance FormulaLet A = ( ─ 2, 5) and B = ( 4, 1). Find the midpoint, C, of AB. Then use the Distance Formula to verify that AC = CB

Solution: Use the Midpoint Formula : C = ( x1 + x2 , y1 + y2 ) 2 2

= ( ─ 2 + 4 , 5 + 1 ) = ( 1 , 3 ) 2 2

AC = (1 – (– 2)2 + ( 3 – 5 )2 = 32 + (– 2)2 = 13

To find AC and CB, use the Distance Formula:

CB = (4 – 1)2 + ( 1 – 3 )2 = 32 + (– 2)2 = 13

Thus, AC = CB

•

•

•A ( ─2, 5)

C ( 1, 3)

B ( 4, 1)

2.32.3

14

Solution Using a Segment BisectorSolution Using a Segment Bisector

●●

●●

●●

●● ●●EE

SS

FFPP

RRRS bisects EF at point Pa.If EF = 12, find PFb.If EP = 4, find EFc.If EP = 4x – 3 and PF = 2x + 15, find EF

SOLUTIONa.PF = ½ EF = ½ (12) = 6b.EF = 2EP = 2 (4) = 8c.Since EP = PF

4x – 3 = 2x + 15 + 3 + 34x = 2x + 18- 2x -2x2x = 182x = 182 2 x = 9

EP = 4x – 3 = 4 (9) – 3 = 36 – 3 = 33

EF = 2 (EP ) = 2 ( 33 ) = 66

2.32.3

The IF … THEN … Sentence StructureThe IF … THEN … Sentence Structure

Consider the statement, “ If I graduate from high school with an average greater than 90, THEN my parents will buy me a car.” IF means “ condition to be met”THEN means “consequence when it does”

THEOREMS in geometry are usually expressed as conditional states in IF – THEN form

After a theorem is proved, the THEN statement is applied in any future After a theorem is proved, the THEN statement is applied in any future proof whenever the IF statement is true.proof whenever the IF statement is true.

Before a theorem is proved, the IF statement is what we know, the THEN Before a theorem is proved, the IF statement is what we know, the THEN statement is what we need to prove.statement is what we need to prove.

For example: “ If a figure is a rectangle, then its diagonals have the same length”,So, in future proofs, whenever you have equal length diagonals, you can assume that the figure is a rectangle.

If a figure is a rectangle, then its diagonals have the same length GIVEN TO BE PROVEN

15

2.42.4

16

Using Conditional Statements

Conditional Statements are the same as IF –THEN statements ,

For example, “If you study at least 3 hours, then you will pass the test.”

A Conditional Statement has 2 parts:1.Hypothesis, denoted by p2.Conclusion, denoted by q

Conditional Statement is written, “ If p, then q” or p q

A Converse Statement is a Conditional Statement reversed ,

For example, the converse of p q is q p

NOTE: a Conditional Statement may be true or false, so you must prove Conditional Statements. You must prove Converse Statements as well.

To prove a statement TRUE , you must present an argument that works for all possible cases.To prove a statement FALSE , you only need to present 1 example to the contrary.

2.42.4

17

Example 1: Conditional Statements and Converses

Decide whether the statement and its converse are true.

a.If m < A = 30o, then < A is acute

b.If m < A = 90o, then < A is right angle

c.If < A = obtuse, then the m < A = 120o

Solution:

a.The statement is true because 30o < 90 o , but the converse “ if <A is acute, then m <A = 30 o “ is false [ some acute angles do not measure 30 o ]

b.Both the statement and the converse are true

c.The statement is false [some obtuse angles have measures that are not 120 o , but the converse is true ]

2.42.4

18

Biconditional Statements

Biconditional Statement is when “ p if and only if q “ or p qWhich is the same as writing both a conditional statement ( p q ) and its converse ( q p ) at the same time.

An example of a Biconditional Statement is “ An angle is a right angle if and only if it measures 90 degrees.” To be a valid biconditional statement it must be true both ways.

2.42.4

19

Translating Conditional StatementsTranslating Conditional Statements

Translate the statement to IF-THEN form.

a.The defendant was in Dallas only on Saturdays

b.Court begins only if it is 10 AM

Solution:

a.A Venn diagram as shown can help translate the statement. In the diagram, the days on which the defendant was in Dallas is a subset of “Saturdays.” In the IF-THEN form, the statement can be written as

If the defendant was in Dallas, then it was Saturday.

b. In general, the statement “ p only If q” is equivalent to “if p, then q”In the IF-THEN form, the statement can be written as

If court begins, then it is 10 AM

Days in Days in which the which the defendant defendant

was in Dallaswas in Dallas

SaturdaysSaturdays

2.42.4

20

Point, Line and Plane Postulates

Postulate 5 Through any 2 distinct points there exists exactly one line

Postulate 6 A line contains at least 2 points

Postulate 7 Through any 3 non-collinear points there exists exactly 1 plane

Postulate 8 A plane contains at least 3 non-collinear points

Postulate 9 If 2 distinct points lie in a plane, then the line containing them lies in the plane

Postulate 10 If 2 distinct planes intersect, then their intersection is a line

2.42.4

Example of Logic and Reasoning to a ProofExample of Logic and Reasoning to a ProofAssume the following two postulates are true:1.All last names that have 7 letters with no vowels are the names of Martians2.All Martians are 3 feet tallProve that Mr. Xhzftlr is 3 feet tall.

PROOF:

Statements Reasons

1. The name is Mr. Xhzftlr Given

2. Mr. Xhzftlr is a MartianAll last names that have 7 letters with no vowels are names of Martians.(See Postulate 1 )

3. Mr. Xhzftlr is 3 feet tall All Martians are 3 feet tall (See Postulate 2 )

Notice that each statement has a corresponding justification21

2.52.5

22

Using Properties from Algebra

Properties of Equality Let a, b, and c be real numbers

Addition Property If a = b, then a + c = b + c

Subtraction Property If a = b, then a – c = b – c

Multiplication Property If a = b, then ac = bc

Division Property If a = b and c ≠ 0, then a c = b c

Reflexive Property For any real number a, a = a

Symmetric Property If a = b, then b = a

Transitive Property If a = b and b = c, then a = c

Substitution Property If a = b, then a may be substituted for b in any equation or expression

2.52.5

23

Using Properties of Congruence

Properties of Congruence

Reflexive Property Any geometric object is congruent to itself

Symmetric Property If one geometric object is congruent to a second, then the second is congruent to the first

Transitive Property If one geometric object is congruent to a second, and the second is congruent to a third, then the first object is congruent to the third

2.52.5

To recognize these properties

Reflexive Property a = a One entity compared

Symmetric Property If a = b, then b = a Two entities compared

Transitive Property If a = b and b = c, then a = c Three entities compared

24

Example of Properties of Length and Measure

Reflexive Property

Segments For any line segment AB, AB = AB

Angles For any angle A, m < A = m < A

Symmetric Property

Segments If AB = CD, then CD = AB

Angles If m < A = m < B, then m < B = m < A

Transitive Property

Segments If AB = CD and CD = EF, the AB = EF

Angles If m < A = m < B and m < B = m < C, then m < A = m < C

2.52.5

INDUCTIVE Versus DEDUCTIVE ReasoningINDUCTIVE Versus DEDUCTIVE Reasoning

String of Odd Integers Sum

1 + 3 4

1 + 3 + 5 9

1 + 3 + 5 + 7 16

1 + 3 + 5 + 7 + 9 25

Consider the result of accumulating consecutive odd integers beginning with 1.

Inductive Reasoning Deductive Reasoning

Do you notice a pattern? It appears that the sum of consecutive odd integers, beginning with 1, will always be a perfect square. The product of the same #.

If, on the basis of this evidence, we now conclude that this relationship will always be true, no matter how many terms are added. Inductive Reasoning involves examining a few examples, observing a pattern, and then assuming that the pattern will never end.

Deductive Reasoning may be considered to be the opposite of Inductive Reasoning. Rather than begin with a few specific instances as is common with inductive processes, deductive reasoning uses accepted facts to reason in a step-by-step fashion until a desired conclusion is reached. 25

2.62.6

26

Two angles are VERTICAL ANGLES if their sides form two pairs of opposite rays.

Two adjacent angles are a LINEAR PAIR if their non-common sides are opposite rays

Two angles are COMPLEMENTARY if the sum of their measures is 90 o . Each angle is the complement of the other

Two angles are SUPPLEMENTARY if the sum of their measures is 180 o . Each angle is the supplement of the other

34

56 6

34

5

21

12

Linear Pair Postulate: If 2 angles form a linear pair, then they are supplementary, the sum of their measures = 180 o

27

Supplementary and Complementary Angle PairsSupplementary and Complementary Angle PairsSupplementary and Complementary Angle PairsSupplementary and Complementary Angle Pairs

Example 3.1 In triangle ABC, angle A is complementary to angle B. Find the measures of angles A and B.

SOLUTION2 x + 3 x = 90 5 x = 90 x = 18

AA

BB

CC(2x) (2x) °°

(3x) (3x) °°

m < A = 2 x = 2 ( 18 ) = 36m < A = 3 x = 3 ( 18 ) = 54

Example 3.2 The measures of an angle and its supplement are in the ratio of 1 : 8. Find the measure of the angle.

SOLUTION [method 1 ]Let x = measure of angle, then 180 – x = measure of supplement of < x = 1180 – x 8 180 – x = 8 x

180 = 9 x 20 = x

SOLUTION [method 2 ]Let x = measure of angle, then 8 x = measure of supplement of <x + 8 x = 180 9 x = 180 x = 20

2.62.6

28

Adjacent Angle PairsAdjacent Angle PairsAdjacent Angle PairsAdjacent Angle Pairs

THEOREM: If the exterior sides of a pair of adjacent angles form a straight line, then the angles are supplementary.

1 2A BC

D

AC and CB are the exterior sides of angles 1 and 2.

150 ° 30 °

REMEMBER: Supplementary angles do NOT have to be adjacent.

2.62.6

29

Adjacent Angle PairsAdjacent Angle PairsAdjacent Angle PairsAdjacent Angle Pairs

Adjacent means “next to” whereas adjacent angles have same vertex, share a common side and have NO interior points in common. Only one of the following are adjacent <‘s

A

A

A

A

11

11

22

2 2

These angles are adjacent because:

These angles do NOT have the same vertex

These angles do have the interior points in common

These angles do NOT share a common side

same vertexcommon sideNO interior points in common

••

2.62.6

30

Theorem: Vertical Angles are CongruentTheorem: Vertical Angles are CongruentTheorem: Vertical Angles are CongruentTheorem: Vertical Angles are Congruent

a. Find the value of xb. Find the measures of angles AEC, DEB, DEA and BEC

( 3 x – 18 ) ° ( 2 x + 5 ) °

a. 3 x – 18 = 2 x + 5 – 2 x + 18 – 2 x + 18 1 x = + 23

b. m < AEC = m < DEB = 3 x – 18 = 3 (23 ) – 18 x = 51Since angles AEC and DEA are supplementarym < DEA = 180 – 51 = 129m < DEA = m < BEC = 129

A

C

E

D

B

2.62.6

31

Theorem: If 2 angles are congruent & supplementary, Theorem: If 2 angles are congruent & supplementary, then each is a right anglethen each is a right angle

Theorem: If 2 angles are congruent & supplementary, Theorem: If 2 angles are congruent & supplementary, then each is a right anglethen each is a right angle

A

CE

D

B

Theorem: If 2 lines intersect to form congruent adjacent Theorem: If 2 lines intersect to form congruent adjacent angles, then the lines are perpendicularangles, then the lines are perpendicular

Theorem: If 2 lines intersect to form congruent adjacent Theorem: If 2 lines intersect to form congruent adjacent angles, then the lines are perpendicularangles, then the lines are perpendicular

Theo

rem

: All

right

ang

les

are

Theo

rem

: All

right

ang

les

are

cong

ruen

tco

ngru

ent

Theo

rem

: All

right

ang

les

are

Theo

rem

: All

right

ang

les

are

cong

ruen

tco

ngru

ent

Theorem: Perpendicular lines

Theorem: Perpendicular lines

intersect to form 4 right angles

intersect to form 4 right angles

Theorem: Perpendicular lines

Theorem: Perpendicular lines

intersect to form 4 right angles

intersect to form 4 right angles

32

Examples of Properties used in Proofs

Given: m < 1 + m < 2 = 90

m < 2 = m < 3

Conclusion: m < 1 + m < 3 = 90

Reason:

12

3

Given: < 1 < 2

< 2 < 3

Conclusion: < 1 < 3

Reason:

Ex 1

Ex 2

1

2

3

NOTE: Since we may only substitute equals in equations, we do not have a Substitution Property of Congruence.

Substitution Property

Transitive Property

2.52.5

33

Examples of Properties used in Proofs

Given: m < 1 = m < 4

m < 3 = m < 5

m < 4 + m < 2 + m < 5 = 180

Conclusion: m < 1 + m < 2 + m < 3 = 180

Reason:

1

2

3

Given: RS = SM

TW = SM

Conclusion: RS = TW

Reason:

Ex 3

Ex 4

T

MR

54

S

W

Transitive Property

Transitive or Substitution Property

2.52.5

34

Examples of Properties used in Proofs

Given:C is the midpoint of AD

AC = CE

Conclusion: CD = CE

Reason:

Ex 5

DC

B

A

ETransitive Property

AC = CD

2.52.5

35

Using the ADDITION PROPERTYUsing the ADDITION PROPERTY

Given: AB = AC + BD = + CE

Conclusion: AB + BD = AC + CE

AD = AE

7 = 7

5

B C

D

2 2

5

E

A

Given: m < JXK = m < MXL + m < KXL = + m < KXL

Conclusion: m < JXL = m < KXM

90 = 90

J

K L

MX

20 o

70 o 70 o

2.52.5

36

Using the SUBTRACTION PROPERTYUsing the SUBTRACTION PROPERTY

Given: VI = NE – EI = – EI

Conclusion: VI + EI = NE + EI

VE = NI

3 = 3

B CP60 o

DAGiven: m < BAD = m < DCB– m < PAD = – m < BCQ

Conclusion: m < BAP = m < DCQ

40 = 40

VV

Q

60 o

100 o 100 o

EE II NN•• •••• ••4433 33

77

77

2.52.5

37

Using the MULTIPLICATION PROPERTYUsing the MULTIPLICATION PROPERTY

BB

CC

TT

AA

Given: AB = CB AR = ½ AB CT = ½ CB

Conclusion: AR = CT WHY ?

Reasoning: RR

Since we are multiplying equals (AB = CB ) by the same number ( ½ ), their products must be equal:½ AB = ½ CBBy substitution, AR = CT

This chain of reasoning, in which the multiplying factor is ½ , is used so often that

we give it a special name : “ halves of equals are equalhalves of equals are equal.”

2.52.5

38

The Two-Column Proof FormatThe Two-Column Proof Format

A Proof in Geometry usually includes these four elements

1) A labeled diagram

2. Given: The set of facts that you can use

3. Prove: What you need to show

4. Proof: Step-by-step reasoning that leads from what is “Given” to what you must “Prove”

2.52.5

39

Given: m < AOC = 50 ° , m < 1 = 25 °

PROVE: OB bisects < AOC

Statements Reasons

m < 1 + m < 2 = m < AOC Angle Addition Postulate

25 ° + m < 2 = 50 ° Substitution

m < 2 = 25 ° Subtraction Property of Equality

Writing an Argument ( Writing an Argument ( Making a ProofMaking a Proof))

21

O

C

B

A•

•

•

•

2.52.5

40

m

π