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2540082550LESSON4-600LESSON4-6Practice B443293513081000Triangle Congruence: CPCTC1.Heike Dreschler set the Woman’s World Junior Record for the long jump in 1983. She jumped about 23.4 feet. The diagram shows two triangles and a pond. Explain whether Heike could have jumped the pond along path BA or along path CA. 3631565118745Reflexive Property ASACPCTC GivenKM KM LKM JMKParallel lines cut by a transversal form congruent alt. int. angles.020000Reflexive Property ASACPCTC GivenKM KM LKM JMKParallel lines cut by a transversal form congruent alt. int. angles.235648511874500Write a proof.2.Given: L J, Prove: LKM JMKStatement Reason1.) 2.) 2.) Given 3.) 4.) 4.) 5.) 6.) 6.) 7.3; 5?; 4?; 3?; 5?; 48.SSS9.CPCTCPractice B1.Possible answer: Because DCE BCA by the Vertical Thm. the triangles are congruent by ASA, and each side in ABC has the same length as its corresponding side in EDC. Heike could jump about 23 ft. The distance along path BA is 20 ft because BA corresponds with DE, so Heike could have jumped this distance. The distance along path CA is 25 ft because CA corresponds with CE, so Heike could not have jumped this distance.2.3.StatementsReasons1. FGHI is a rectangle.1. Given2. FIH and GHI are right angles.2. Def. of rectangle3. FIH GHI 3. Rt. Thm.4. 4. Reflex. Prop. of 5. FIH GHI5. SAS6. 6. CPCTC7. FH GI7. Def. of segs.Practice C1.Possible answer: From the definition of a parallelogram, is congruent to and is parallel to . By the Alternate Interior Angles Theorem, BAC is congruent to DCA and CDB is congruent to ABD. Therefore ABE is congruent to CDE by ASA. By CPCTC, is congruent to and is congruent to . Congruent segments have equal lengths, so the diagonals bisect each other.2.Possible answer: From the definition of a rhombus, is congruent to , is congruent to , and is parallel to . By Alternate Interior Angles Theorem, GFH is congruent to IHF and FGI is congruent to HIG. Therefore FGJ is congruent to HIJ by ASA. By CPCTC, is congruent to and is congruent to . So FJI is congruent to GHJ by SSS. But HIJ is also congruent to FIJ by SSS. And so all four triangles are congruent by the Transitive Property of Congruence. By CPCTC and the Segment Addition Postulate, is congruent to . By CPCTC and the Linear Pair Theorem, FJI, GJF, HJG, and IJH are right angles. So and are perpendicular. By CPCTC, GFH, IFH, GHF, and IHF are congruent, so bisects IFG and IHG. Similar reasoning shows that bisects FGH and FIH.3.The diagonals of a rectangle bisect each other.4.The diagonals of a square are congruent perpendicular bisectors that bisect the vertex angles of the square.5.The diagonals are congruent.Reteach1.-6667569215002.StatementsReasons1. UXW and UVW are rt. s.1. Given2. 2. a. Given ................
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