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Reasoning with Properties of Algebra & Proving Statements About Segments CCSS: G-CO.12 A proof is an argument that uses logic, definitions, properties, and previously proven statements to show that a conclusion is true. An important part of writing a proof is giving justifications to show that every step is valid. You learned in Chapter 1 that segments with equal lengths are congruent and that angles with equal measures are congruent. So the Reflexive, Symmetric, and Transitive Properties of Equality have corresponding properties of congruence. Theorem • A true statement that follows as a result of other true statements. • All theorems MUST be proved! 2-Column Proof • Numbered statements and corresponding reasons in a logical order organized into 2 columns. statements reasons 1. 1. 2. 2. 3. 3. etc. Properties of Segment and Angle Congruence Properties of Segment and Angle Equality • *We use these to prove properties of congruence Theorem 2.1- Properties of Segment Congruence • Segment congruence is reflexive, symmetric, & transitive. For any AB, AB AB. If AB BC and BC CD, then AB CD. If AB BC, then BC AB. Proof of symmetric part of thm. 2.1 Statements 1. AB BC 2. AB = BC 3. BC = AB 4. BC AB Reasons 1. 2. 3. 4. Given Defn. of congruent segs. Symmetric prop of = Defn. of congruent segs. Paragraph Proof • Same argument as a 2-column proof, but each step is written as a sentence; therefore forming a paragraph. P X Y Q • You are given that line segment PQ is congruent with line segment XY. By the definition of congruent segments, PQ=XY. By the symmetric property of equality XY = PQ. Therefore, by the definition of congruent segments, it follows that line segment XY congruent to line segment PQ. Ex: Given: LK = 5 JK = 5 JK ≅ JL Prove: LK≅ JL Statements 1. 2. 3. LK = JK 4. LK ≅ JK 5. JK ≅ JL 6. K Reasons 1. Given 2. Given 3. Trans. Prop of = 4. 5. Given 6. Trans. Prop of congruence Ex: Given: Q is the midpoint of PR. PR Prove: PQ and QR = 2 1. 2. 3. 4. 5. 6. Statements Q is midpt of PR PQ=QR PQ+QR=PR QR+QR=PR 2QR=PR QR= PR 2 PR 7. PQ= 2 1. 2. 3. 4. 5. 6. Reasons Given Defn. of midpt Seg + post Subst. prop of = Simplify Division prop of = 7. Subst. prop Classwork • Page 105: problems 6-11 all • We’ll go over the answers after