G E O M E T R Y Chapter 2: Measure & Congruence

1. GEOMETRY Chapter 2: Measure & Congruence

2. 2.1 Measuring Segments and Angles To measure segments we use a ruler. In this example, the length or measure of line segment AB is 2 inches We write this as mAB We cannot write AB = 2 Since the number of points between A and B are infinite and NOT limited to 2. ● ● A B 1 2 3 4 5 6 INCHES

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

4. 2.1 Classifying Angles a ° a ° a °

5. 2.1 Betweenness of Points and Rays Definition of Betweeness: Point C is between points A and B if both the following conditions are met: Points A, C, and B are three different collinear points AB = AC + CB 2 3 ● ● ● A C B Example: Point C is between A and B. If AC = 2 and CB = 3, find AB A If m < AOP = 40 and m < POB = 20 then m < AOB = 60. This logical relationship is called: ● P 60 ° ● Angle Addition Postulate 40 ° 20 ° ● If ray OP is in the interior of <AOB, then m < AOP = m < AOP + m < POB O ● B

6. 2.1 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

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

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

9. 2.2 Example 3 of Angle Addition Postulate 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. ● G 90 ° ? B 130 ° ● ● C Interior of < ABC A ● 140 ° REASON: BG is not in the interior of < ABC, thus violating the assumption (or hypothesis ) of the Angle Addition Postulate.

10. 2.3 Congruence Figures that have the same shape and size are said to be CONGRUENT. 4 4 Same size but not the same shape 4 4 4 4 4 4 If line segments have the same length, they are congruent. Notation: AB RS Same shape but not the same size 2 If 2 angles have the same measure , they are congruent. m < ABC m < DEF 2 2 2 REMEMBER: Figures are congruent only if they agree in all their dimensions.

11. 2.3 Midpoint and Bisector 3 3 • DEFINITION OF MIDPOINT • Point M is the midpoint of AB if • M is between A and B • AM = MB ● ● ● A M B X ● M 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. ● ● ● A B ● Y

12. 1.4 The Midpoint Formula • A ( ─2, 5) • The midpoint of the line segment joining A (x1, y1) and B (x2, y2) is as follows: C ( 1, 3) • B ( 4, 1) Each coordinate of M is the mean of the corresponding coordinates of A and B.

13. Let A = ( x1 , y1) and B = ( x2 , y2 ) be points in a coordinate plane. The distance between A and B is 2.3 The Distance Formula AB = ( x2─ x1 ) 2 + ( y2─ y1 ) 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 = ( x1 + x2, y1 + y2 ) 2 2 = ( ─ 2 + 4 , 5 + 1 ) = ( 1 , 3 ) 2 2 To find AC and CB, use the Distance Formula: A ( ─2, 5) AC = (1 – (– 2)2 + ( 3 – 5 )2 = 32 + (– 2)2 = 13 • C ( 1, 3) • CB = (4 – 1)2 + ( 1 – 3 )2 = 32 + (– 2)2 = 13 • B ( 4, 1) Thus, AC = CB

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

15. 2.4 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. 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. Before a theorem is proved, the IF statement is what we know, the THEN statement is what we need to prove. If a figure is a rectangle, then its diagonals have the same length GIVEN TO BE PROVEN

16. 2.4 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: • Hypothesis, denoted by p • Conclusion, denoted by q • Conditional Statement is written, “ If p, then q”or p q • A Converse Statement is aConditional 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.

17. 2.4 Example 1: Conditional Statements and Converses • Decide whether the statement and its converse are true. • If m < A = 30o, then < A is acute • If m < A = 90o, then < A is right angle • If < A = obtuse, then the m < A = 120o • Solution: • 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 ] • Both the statement and the converse are true • The statement is false [some obtuse angles have measures that are not 120 o , but the converse is true ]

18. 2.4 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.

19. 2.4 Translating Conditional Statements • Translate the statement to IF-THEN form. • The defendant was in Dallas only on Saturdays • Court begins only if it is 10 AM • Solution: • 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 Saturdays Days in which the defendant was in Dallas

20. 2.4 Point, Line and Plane Postulates

21. 2.5 Example of Logic and Reasoning to a Proof • Assume the following two postulates are true: • All last names that have 7 letters with no vowels are the names of Martians • All Martians are 3 feet tall • Prove that Mr. Xhzftlr is 3 feet tall. Notice that each statement has a corresponding justification

22. 2.5 Using Properties from Algebra

23. 2.5 Using Properties of Congruence

24. 2.5 Example of Properties of Length and Measure

25. 2.6 INDUCTIVE Versus DEDUCTIVE Reasoning Consider the result of accumulating consecutive odd integers beginning with 1.

26. 3 3 4 6 4 6 5 5 2 1 1 2 Linear Pair Postulate: If 2 angles form a linear pair, then they are supplementary, the sum of their measures = 180 o

27. 2.6 Supplementary and Complementary Angle Pairs B 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 (3x) ° (2x) ° m < A = 2 x = 2 ( 18 ) = 36 m < A = 3 x = 3 ( 18 ) = 54 A C 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 SOLUTION [method 2 ]Let x = measure of angle, then 8 x = measure of supplement of < x + 8 x = 180 9 x = 180 x = 20 180 – x = 8 x 180 = 9 x 20 = x

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

29. 2.6 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 1 1 2 A • 2 These angles are adjacent because: A same vertex common side NO interior points in common These angles do NOT have the same vertex 1 1 2 2 A A These angles dohavethe interior points in common These angles do NOT share a common side

30. 2.6 Theorem: Vertical Angles are Congruent • Find the value of x • Find the measures of angles AEC, DEB, DEA and BEC A D • 3 x – 18 = 2 x + 5 • – 2 x + 18 – 2 x + 18 • 1 x = + 23 ( 3 x – 18 ) ° ( 2 x + 5 ) ° E C • 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 B

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

32. 2.5 Examples of Properties used in Proofs Ex 1 3 1 2 Substitution Property Ex 2 1 3 2 Transitive Property NOTE: Since we may only substitute equals in equations, we do not have a Substitution Property of Congruence.

33. 2.5 Examples of Properties used in Proofs Ex 3 4 5 2 1 3 Transitive Property Ex 4 S T R W M Transitive or Substitution Property

34. 2.5 Examples of Properties used in Proofs Ex 5 B C A D AC = CD Transitive Property E

35. 2.5 Using the ADDITION PROPERTY A 2 2 B C 5 5 D E K L 20 o J 70 o 70 o X M

36. 2.5 Using the SUBTRACTION PROPERTY Q D A 60 o 100 o 100 o 60 o B P C 7 3 3 4 • • • • V N E I 7

37. 2.5 Using the MULTIPLICATION PROPERTY B R T 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 A C 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.”

38. 2.5 The Two-Column Proof Format

39. 2.5 Writing an Argument ( Making a Proof) C • B • 2 • 1 A • O

40. m  π 