1. 1

2. 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 write this as mAB We cannot write AB = 2 INCHES AB 2 2.1 Since the number of points between A and B are infinite and NOT limited to 2.

3. Measuring Segments and Angles To measure angles we use a protractor. ● ● ● ● ● ● ● APF Z B W S For example, the measure of angle ZPA is 50 ° written as m< ZPA = 50 ° 3 2.1

4. Classifying 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 ° 4 2.1

5. Betweenness 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 points 2.AB = AC + CB ●●● ABC 23 Example: Point C is between A and B. If AC = 2 and CB = 3, find AB 60 ° 20 ° 40 ° A B P O ● ● ● ● If m < AOP = 40 and m < POB = 20 then m < AOB = 60. This logical relationship is called: Angle Addition Postulate If ray OP is in the interior of <AOB, then m < AOP = m < AOP + m < POB 5 2.1

6. INITIAL 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 Two points determine a line ● ● A B POSTULATE 1.2 Three non-collinear points determine a plane A ● B ● ● C 6 2.1

7. BG lies in the interior of < ABC. If m < ABG = 30 and m < GBC = 20, then find m < ABC 20 ° 30 ° A C G B ● ● ● ● 7 Example 1 of Angle Addition Postulate ? 2.2

8. KM lies in the interior of < JKL. If m < JKL = 50 and m < MKL = 20, then find m < ABC 20 ° 50 ° J L M K ● ● ● ● The Angle Addition Postulate may also be expressed as m < JKM = m < JKL - m < MKL 8 Example 2 of Angle Addition Postulate ? 2.2

9. Here the Angle Addition Postulate is contradicted, the measure of the largest angle is NOT equal to the sum of the measures of the 2 smaller angles. 130 ° 90 ° G C A B ● ● ● ● 9 Example 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 2.2

10. 10 Congruence CONGRUENT. Figures that have the same shape and size are said to be CONGRUENT. REMEMBER: Figures are congruent only if they agree in all their dimensions. 4 44 4 4 4 4 4 Same 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 2.3

11. 11 Midpoint and Bisector ● ●● B AM ●●● AB M X Y ● ● DEFINITION OF MIDPOINT Point M is the midpoint of AB if 1.M is between A and B 2.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 2.3

12. 12 The midpoint of the line segment joining A (x 1, y 1 ) and B (x 2, y 2 ) is as follows: A ( ─ 2, 5) C ( 1, 3) B ( 4, 1) M = ( x 1 + x 2, y1 + y2y1 + y2 ) 2 2 Each coordinate of M is the mean of the corresponding coordinates of A and B. M = ( ─ 2 + 4, 5 + 15 + 1 ) 2 2 M = ( 2, 6 ) 2 2 The Midpoint Formula 1.4

13. 13 The Distance Formula Let A = ( x 1, y 1 ) and B = ( x 2, y 2 ) be points in a coordinate plane. The distance between A and B is AB = ( x 2 ─ x 1 ) 2 + ( y 2 ─ y 1 ) 2 Example 1 Using the Distance Formula Let 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 = ( x 1 + x 2, y 1 + y 2 ) 2 2 = ( ─ 2 + 4, 5 + 1 ) = ( 1, 3 ) 2 2 AC = (1 – (– 2) 2 + ( 3 – 5 ) 2 = 3 2 + (– 2) 2 = 13 To find AC and CB, use the Distance Formula: CB = (4 – 1) 2 + ( 1 – 3 ) 2 = 3 2 + (– 2) 2 = 13 Thus, AC = CB A ( ─ 2, 5) C ( 1, 3) B ( 4, 1) 2.3

14. 14 Solution Using a Segment Bisector ● ● ● ●●E S F P R RS bisects EF at point P a.If EF = 12, find PF b.If EP = 4, find EF c.If EP = 4x – 3 and PF = 2x + 15, find EF SOLUTION a.PF = ½ EF = ½ (12) = 6 b.EF = 2EP = 2 (4) = 8 c.Since EP = PF 4x – 3 = 2x + 15 + 3 + 3 4x = 2x + 18 - 2x -2x 2x = 18 2 x = 9 EP = 4x – 3 = 4 (9) – 3 = 36 – 3 = 33 EF = 2 (EP ) = 2 ( 33 ) = 66 2.3

15. The 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 proof whenever the IF statement is true. Before a theorem is proved, the IF statement is what we know, the THEN 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 2.4

16. 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 p 2.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. 2.4

17. 17 Example 1: Conditional Statements and Converses Decide whether the statement and its converse are true. a.If m < A = 30 o, then < A is acute b.If m < A = 90 o, then < A is right angle c.If < A = obtuse, then the m < A = 120 o Solution: a.The statement is true because 30 o < 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 ] 2.4

18. 18 Biconditional Statements Biconditional Statement is when “ p if and only if q “ or p q Which 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. 2.4

19. 19 Translating 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 which the defendant was in Dallas Saturdays 2.4

20. 20 Point, Line and Plane Postulates Postulate 5Through any 2 distinct points there exists exactly one line Postulate 6A line contains at least 2 points Postulate 7Through any 3 non-collinear points there exists exactly 1 plane Postulate 8A plane contains at least 3 non-collinear points Postulate 9If 2 distinct points lie in a plane, then the line containing them lies in the plane Postulate 10If 2 distinct planes intersect, then their intersection is a line 2.4

21. Example of Logic and Reasoning to a Proof Assume the following two postulates are true: 1.All last names that have 7 letters with no vowels are the names of Martians 2.All Martians are 3 feet tall Prove that Mr. Xhzftlr is 3 feet tall. PROOF: StatementsReasons 1. The name is Mr. XhzftlrGiven 2. Mr. Xhzftlr is a Martian All 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 justification 21 2.5

22. 22 Using Properties from Algebra Properties of EqualityLet a, b, and c be real numbers Addition PropertyIf a = b, then a + c = b + c Subtraction PropertyIf a = b, then a – c = b – c Multiplication PropertyIf a = b, then ac = bc Division Property   If a = b and c ≠ 0, then a  c = b  c Reflexive PropertyFor any real number a, a = a Symmetric PropertyIf a = b, then b = a Transitive PropertyIf a = b and b = c, then a = c Substitution PropertyIf a = b, then a may be substituted for b in any equation or expression 2.5

23. 23 Using Properties of Congruence Properties of Congruence Reflexive PropertyAny geometric object is congruent to itself Symmetric PropertyIf one geometric object is congruent to a second, then the second is congruent to the first Transitive PropertyIf 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 2.5 To recognize these properties Reflexive Property a = aOne entity compared Symmetric PropertyIf a = b, then b = aTwo entities compared Transitive PropertyIf a = b and b = c, then a = cThree entities compared

24. 24 Example of Properties of Length and Measure Reflexive Property SegmentsFor any line segment AB, AB = AB AnglesFor any angle A, m < A = m < A Symmetric Property SegmentsIf AB = CD, then CD = AB AnglesIf m < A = m < B, then m < B = m < A Transitive Property SegmentsIf AB = CD and CD = EF, the AB = EF AnglesIf m < A = m < B and m < B = m < C, then m < A = m < C 2.5

25. INDUCTIVE Versus DEDUCTIVE Reasoning String of Odd IntegersSum 1 + 34 1 + 3 + 59 1 + 3 + 5 + 716 1 + 3 + 5 + 7 + 925 Consider the result of accumulating consecutive odd integers beginning with 1. Inductive ReasoningDeductive 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 2.6

26. 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 3 4 5 66 3 4 5 2 1 12 Linear Pair Postulate: If 2 angles form a linear pair, then they are supplementary, the sum of their measures = 180 o

27. 27 Supplementary 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. SOLUTION 2 x + 3 x = 90 5 x = 90 x = 18 A B C (2x) ° (3x) ° m < A = 2 x = 2 ( 18 ) = 36 m < 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 = 1 180 – 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 2.6

28. 28 Adjacent Angle Pairs THEOREM: If the exterior sides of a pair of adjacent angles form a straight line, then the angles are supplementary. 12 ABC D AC and CB are the exterior sides of angles 1 and 2. 150 ° 30 ° REMEMBER: Supplementary angles do NOT have to be adjacent. 2.6

29. 29 Adjacent 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 1 1 1 1 2 2 22 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 vertex common side NO interior points in common 2.6

30. 30 Theorem: Vertical Angles are Congruent a.Find the value of x b.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 = 51 Since angles AEC and DEA are supplementary m < DEA = 180 – 51 = 129 m < DEA = m < BEC = 129 A C E D B 2.6

31. 31 Theorem: If 2 angles are congruent & supplementary, then each is a right angle A C E D B Theorem: If 2 lines intersect to form congruent adjacent angles, then the lines are perpendicular Theorem: All right angles are congruent Theorem: Perpendicular lines intersect to form 4 right angles

32. 32 Examples of Properties used in Proofs Given:m < 1 + m < 2 = 90 m < 2 = m < 3 Conclusion:m < 1 + m < 3 = 90 Reason: 1 2 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 2.5

33. 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 M R 5 4 S W Transitive Property Transitive or Substitution Property 2.5

34. 34 Examples of Properties used in Proofs Given: C is the midpoint of AD AC = CE Conclusion:CD = CE Reason: Ex 5 D C B A E Transitive Property AC = CD 2.5

35. 35 Using the ADDITION PROPERTY Given: AB = AC + BD = + CE Conclusion: AB + BD = AC + CE AD = AE 7 = 7 5 B C D 22 5 E A Given: m < JXK = m < MXL + m < KXL = + m < KXL Conclusion: m < JXL = m < KXM 90 = 90 J K L M X 20 o 70 o 2.5

36. 36 Using the SUBTRACTION PROPERTY Given: VI = NE – EI = – EI Conclusion: VI + EI = NE + EI VE = NI 3 = 3 B CP 60 o D A Given: m < BAD = m < DCB – m < PAD = – m < BCQ Conclusion: m < BAP = m < DCQ 40 = 40 V Q 60 o 100 o EI N 4 33 7 7 2.5

37. 37 Using the MULTIPLICATION PROPERTY B C T A Given: AB = CB AR = ½ AB CT = ½ CB Conclusion: AR = CT WHY ? Reasoning: R Since we are multiplying equals (AB = CB ) by the same number ( ½ ), their products must be equal: ½ AB = ½ CB By substitution, AR = CT halves of equals are equal 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 equal.” 2.5

38. 38 The 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” 2.5

39. 39 Given: m < AOC = 50 °, m < 1 = 25 ° PROVE: OB bisects < AOC StatementsReasons m < 1 + m < 2 = m < AOCAngle Addition Postulate 25 ° + m < 2 = 50 ° Substitution m < 2 = 25 ° Subtraction Property of Equality Writing an Argument ( Making a Proof) 2 1 O C B A 2.5

40. 40 m π  