Provide reasons for this proof. “If a Write a proof for: “If a

EXAMPLE 1 GIVEN: › , intersecting at E See Figure 1.61 PROVE: ⬔AEC is a right angle PROOF Statements Reasons 1. › , intersecting at E 1. Given

1 2.

⬔AEC ⬵ ⬔CEB 2. Perpendicular lines meet to form congruent adjacent angles Definition 2 3. m ⬔AEC = m ⬔CEB 3. If two angles are congruent, their measures are equal 4. ⬔AEB is a straight angle 4. Measure of a straight angle equals 180° and m ⬔AEB = 180° 5. m ⬔AEC + m ⬔CEB = m ⬔AEB 5. Angle-Addition Postulate 4, 5 6. m ⬔AEC + m ⬔CEB = 180° 6. Substitution 3, 6 7. m ⬔AEC + m ⬔AEC = 180° 7. Substitution or 2 · m ⬔AEC = 180° 7 8. m ⬔AEC = 90° 8. Division Property of Equality 8 9. ⬔AEC is a right angle 9. If the measure of an angle is 90°, then the angle is a right angle 쮿 Í CD Í AB Í CD Í AB B A E C D Figure 1.61 RELATIONS The relationship between perpendicular lines suggests the more general, but undefined, mathematical concept of relation. In general, a relation “connects” two elements of an as- sociated set of objects. Table 1.8 provides several examples of the concept of a relation R. TABLE 1.8 Relation R Objects Related Example of Relationship is equal to numbers 2 + 3 = 5 is greater than numbers 7 ⬎ 5 is perpendicular to lines › m is complementary to angles ⬔1 is comp. to ⬔2 is congruent to line segments ⬵ is a brother of people Matt is a brother of Phil CD AB Exs. 1, 2 Reminder Numbers that measure may be equal AB = CD or m⬔1 = m⬔2 whereas geometric figures may be congruent ⬵ or ⬔1 ⬵ ⬔2. CD AB There are three special properties that may exist for a given relation R. Where a, b, and c are objects associated with relation R, the properties consider one object reflexive, two objects in either order symmetric, or three objects transitive. For the properties to Congruence of angles or of line segments is closely tied to equality of angle measures or line segment measures by the definition of congruence. The following list gives some useful properties of the congruence of angles. Reflexive: ⬔1 ⬵ ⬔1; an angle is congruent to itself. Symmetric: If ⬔1 ⬵ ⬔2, then ⬔2 ⬵ ⬔1. Transitive: If ⬔1 ⬵ ⬔2 and ⬔2 ⬵ ⬔3, then ⬔1 ⬵ ⬔3. Any relation such as congruence of angles that has reflexive, symmetric, and transitive properties is known as an equivalence relation. In later chapters, we will see that congruence of triangles and similarity of triangles also have reflexive, symmetric, and transitive properties; these relations are also equivalence relations. Returning to the formulation of a proof, the final example in this section is based on the fact that vertical angles are congruent when two lines intersect. See Figure 1.62. Because there are two pairs of congruent angles, the Prove could be stated Prove: ⬔1 ⬵ ⬔3 and ⬔2 ⬵ ⬔4 Such a conclusion is a conjunction and would be proved if both congruences were es- tablished. For simplicity, the Prove of Example 3 is stated Prove: ⬔2 ⬵ ⬔4 Study this proof of Theorem 1.6.2, noting the order of the statements and reasons. 3 2

4 1

D A C B O Figure 1.62 EXAMPLE 2 Does the relation “is less than” for numbers have a reflexive property? a symmetric property? a transitive property? Solution Because “2 2” is false, there is no reflexive property. “If 2 5, then 5 2” is also false; there is no symmetric property. “If 2 5 and 5 9, then 2 9” is true; there is a transitive property. NOTE: The same results are obtained for choices other than 2, 5, and 9. 쮿 Exs. 3–9 Geometry in Nature An icicle formed from freezing water assumes a vertical path. © Karel Broz ˘Shutterstock If two lines intersect, then the vertical angles formed are congruent. THEOREM 1.6.2 Reflexive property: a Ra 5 = 5; equality of numbers has a reflexive property Symmetric property: If aRb, then bRa. If › m , then m › ; perpendicularity of lines has a symmetric property Transitive property: If aRb and bRc, then aRc. If ⬔1 ⬵ ⬔2 and ⬔2 ⬵ ⬔3, then ⬔1 ⬵ ⬔3; congruence of angles has a transitive property exist, it is necessary that the statements be true for all objects selected from the associated set. These properties are generalized and given examples as follows: EXAMPLE 3 GIVEN: intersects at O See Figure 1.62 on pge 48. PROVE: ⬔2 ⬵ ⬔4 PROOF Statements Reasons 1. intersects at O 1. Given 2. ⬔s AOC and DOB are straight 2. The measure of a straight angle is 180° ⬔s, with m⬔AOC = 180 and m ⬔DOB = 180 3. m ⬔AOC = m ⬔DOB 3. Substitution 4. m ⬔1 + m ⬔4 = m ⬔DOB and 4. Angle-Addition Postulate m ⬔1 + m ⬔2 = m ⬔AOC 5. m ⬔1 + m ⬔4 = m ⬔1 + m ⬔2 5. Substitution 6. m ⬔4 = m ⬔2 6. Subtraction Property of Equality 7. ⬔4 ⬵ ⬔2 7. If two angles are equal in measure, the angles are congruent 8. ⬔2 ⬵ ⬔4 8. Symmetric Property of Congruence of Angles 쮿 Í BD Í AC Í BD Í AC Technology Exploration Use computer software if available. 1. Construct and to intersect at point O. See Figure 1.62. 2. Measure ⬔1, ⬔2, ⬔3, and ⬔4. 3. Show that m⬔1 = m⬔3 and m⬔2 = m⬔4. Í BD Í AC In the preceding proof, there is no need to reorder the congruent angles from state- ment 7 to statement 8 because congruence of angles is symmetric; in the later work, statement 7 will be written to match the Prove statement even if the previous line does not have the same order. The same type of thinking applies to proving lines perpendi- cular or parallel: The order is simply not important CONSTRUCTIONS LEADING TO PERPENDICULAR LINES Construction 2 in Section 1.2 determined not only the midpoint of but also that of the perpendicular bisector of . In many instances, we need the line perpendicular to another line at a point other than the midpoint of a segment. AB AB B A X a B A X b C D B A X c E C D Construction 5 To construct the line perpendicular to a given line at a specified point on the given line. GIVEN: with point X in Figure 1.63a CONSTRUCT: A line , so that ⊥ CONSTRUCTION: Figure 1.63b: Using X as the center, mark off arcs of equal radii on each side of X to intersect at C and D. Figure 1.63c: Now, using C and D as centers, mark off arcs of equal radii with a length greater than XD so that these arcs intersect either above as shown or below . Calling the point of intersection E, draw , which is the desired line; that is, ⊥ . Í AB Í EX Í EX Í AB Í AB Í AB Í EX Í EX Í AB Figure 1.63 Construction 2, which was used to locate the midpoint of a line segment in Section 1.2, is also the method for constructing the perpendicular bisector of a line segment. In Figure 1.64, is the perpendicular bisector of . The following theorem can be proved by methods developed later in this book. RS Í XY The perpendicular bisector of a line segment is unique. THEOREM 1.6.4 Figure 1.64 2 Q N O M P 1 3 Exercises 1.6 In Exercises 1 and 2, supply reasons.

1. Given: ⬔1 ⬵ ⬔3

Prove: ⬔MOP ⬵ ⬔NOQ

2. Given: intersects at

O so that ⬔1 is a right ⬔ Use the figure following Exercise 1. Prove: ⬔2 and ⬔3 are complementary PROOF Statements Reasons 1. intersects at O 1. ? 2. ⬔AOB is a straight ⬔, 2. ? so m ⬔AOB = 180 3. m ⬔1 + m ⬔COB 3. ? = m ⬔AOB 4. m ⬔1 + m ⬔COB = 180 4. ? 5. ⬔1 is a right angle 5. ? 6. m ⬔1 = 90 6. ? 7. 90 + m ⬔COB = 180 7. ? 8. m ⬔COB = 90 8. ? 9. m ⬔2 + m ⬔3 = m ⬔COB 9. ? 10. m ⬔2 + m ⬔3 = 90 10. ? 11. ⬔2 and ⬔3 are 11. ? complementary Í CD Í AB Í CD Í AB PROOF Statements Reasons 1. ⬔1 ⬵ ⬔3 1. ? 2. m ⬔1 = m ⬔3 2. ? 3. m ⬔1 + m ⬔2 = m ⬔MOP 3. ? and m ⬔2 + m ⬔3 = m ⬔NOQ 4. m ⬔1 + m ⬔2 = 4. ? m ⬔2 + m ⬔3 5. m ⬔MOP = m ⬔NOQ 5. ? 6. ⬔MOP ⬵ ⬔NOQ 6. ? B A O C D 2 1 3 Exercise 2 2 1 3 Exercise 3 Exs. 10–14 X M S R Y The theorem that Construction 5 is based on is a consequence of the Protractor Pos- tulate, and we state it without proof. In a plane, there is exactly one line perpendicular to a given line at any point on the line. THEOREM 1.6.3 In Exercises 3 and 4, supply statements.

3. Given: ⬔1 ⬵ ⬔2 and ⬔2 ⬵ ⬔3