LESSON 2-7 Focus on Mathematical Content As you learned in Chapter 1, a segment can be measured, and measures can be used in calculations because they are real numbers. The Ruler Postulate states that the points on any line or line segment can be paired with real numbers so that, given any two points A and B on a line, A corresponds to 0, and B corresponds to a positive real number. That number is the length of the segment. Another postulate states that if point B lies between points A and C on the same line, AB + BC = AC. The converse statement holds true as well. AB + BC = AC The Reflexive, Symmetric, and Transitive Properties of Equality can be used to write proofs about segment congruence. The theorem resulting from the proofs states that congruence of segments is reflexive, symmetric, and transitive. 1. Prove that if AB ≅ CD, then AC ≅ BD Proof: Statements Reasons 1. AB ≅ CD 2. AB = CD 3. BC = BC 4. AB + BC = AC 5. CD + BC = AC 6. CD + BC = BD 7. AC = BD 8. AC ≅ BD 2. Segment Congruence Jamie is designing a badge for her club. The length of the top edge of the badge is equal to the length of the left edge of the badge. The top edge of the badge is congruent to the right edge of the badge, and the right edge of the badge is congruent to the bottom edge of the badge. Prove that the bottom edge of the badge is congruent to the left edge of the badge. Given: WY = YZ YZ ≅ XZ XZ ≅ WX Prove: WX ≅ WY LESSON 2-7 Proof: Statements Reasons 1. WY = YZ YZ ≅ XZ XZ ≅ WX 2. WY ≅ YZ 3. YZ ≅ WX 4. WX ≅ WY DISTANCE Martha and Laura live 1400 meters apart. A library is opened between them and is 500 meters from Martha. How far is the library from Laura? Complete this proof. Given: C is the midpoint of ̅̅̅̅ 𝐵𝐷 and ̅̅̅̅ 𝐴𝐸 . Prove: AB = DE Statements Reasons 1. C is the midpoint of ̅̅̅̅ 𝐵𝐷 and ̅̅̅̅ 𝐴𝐸 . 1. Given 2. BC = CD and _____________ 2. 3. AC = AB + BC, CE = CD + DE 3. 4. AB = AC – BC 4. 5. 5. Substi. Prop. of = 6. DE = CE – CD 6. LESSON 2-8 Focus on Mathematical Content This lesson introduces postulates and theorems about angle relationships. The Protractor Postulate states, The Protractor Postulate states, “Given and a number r between 0 and 180, there is exactly one ray with end-point A, extending on either side of such that the measure of the angle formed is r.” The Angle Addition Postulate states that if R is in the interior of ∠PQS, then m∠PQR + m∠RQS = m∠PQS. The converse is also true. What do the angles of rectangles and squares have in common? ___________________ If a square and a rectangle are adjacent, what is the sum of the measure of any two adjacent angles? ________________________________________________________________________ ________________________________________________________________________ Supplementary and Complementary Angles I. Using a protractor, a construction worker measures that the angle a beam makes with the ceiling is 42°. What is the measure of the angle the beam makes with the wall? ___________________________________________________________________ ___________________________________________________________________ II. At 4 o'clock, the angle between the hour and minute hands of a clock is 120°. When the second hand bisects the angle between the hour and minute hands, what are the measures of the angles between the minute and second hands and between the second and hour hands? ___________________________________________________________________ ___________________________________________________________________ LESSON 2-8 Congruent and Right Angles III. In the figure, ∠1 and ∠4 form a linear pair, and m∠3 + m∠1 = 180. Prove that ∠3 and ∠4 are congruent. Statements Reasons 1. m∠3 + m∠1 = 180; ∠1 and ∠4 form a linear pair. 2. ∠1 and ∠4 are supplementary. 3. ∠3 and ∠1 are supplementary. 4. ∠3 ≅ ∠4 IV. If ∠1 and ∠2 are vertical angles and m∠1 = d − 32 and m∠2 = 175 − 2d, find m∠1 and m∠2. Justify each step. ______________________________________________________________________ ______________________________________________________________________ Statements 1. ∠1 and ∠2 are vertical angles. 2. ∠1 ≅ ∠2 3. 4. 5. 6. m∠1 = m∠2 d − 32 = 175 − 2d 3d = 207 d = 69 7. m∠1 = 37 8. m∠2 = 37 Reasons