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GBK Geometry

GBK Geometry. Jordan Johnson. Today’s plan. Greeting Quiz: Distance Lesson: SSS Homework / Questions Clean-up. Quiz: Distance. Take out a clean sheet of paper you can hand in. Title it “Distance Formula Quiz”

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GBK Geometry

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  1. GBK Geometry Jordan Johnson

  2. Today’s plan • Greeting • Quiz: Distance • Lesson: SSS • Homework / Questions • Clean-up

  3. Quiz: Distance • Take out a clean sheet of paper you can hand in. • Title it “Distance Formula Quiz” • Put your name, today’s date, and your period number at the top-right corner. • Suppose you have a triangle with vertices C(6, –2),R(12, 1), and T(8, 4). • Find the exact length of each side of the triangle.(Simplify the radicals.) • When done, work on one of: • Asg #28 • Circle center puzzle.

  4. Northridge, 1994

  5. Los Angeles, 1994

  6. Real Support

  7. SSS Congruence • Theorem 11 (on p. 164) states: • If the three sides of one triangle are equal to the three sides of another triangle,then the triangles are congruent.

  8. Proving the SSS Congruence Theorem • Strategy for our proof: • Make a copy of DFE, called APC, that shares side AC with ABC.By construction, APC  DFE. • Show that ABC  APC. • Conclude that ABC  DFE by transitivity of “”. (In other words, because if two triangles are congruent to a third triangle, they’re congruent to each other.)

  9. Part I: Duplicate DEF • Measure D with your protractor: • Draw a ray called AX to form XAC such that XAC = D:

  10. Part I: Duplicate DEF (cont.) • Measure DF, and choose a point P on AX so that AP = DF: • Draw segment PC:

  11. Part I: Duplicate DEF (cont.) • APC  DFE by SAS: • AP = DF (by the Ruler Postulate) • PAC = EDF (by the Protractor Postulate) • AC = DE (given)

  12. Part II: Prove ABC  APC (cont.) • Next, draw segment BP and label 1, 2, 3, and 4: • BP divides the figure into two isosceles triangles, PAB and PCB.

  13. Part II: Prove ABC  APC (cont.) • By the isosceles theorems (Thm. 9): • 1 = 2 (because AB = AP) • 3 = 4 (because BC = PC) • By addition, 1 + 3 = 2 + 4.

  14. Part II: Prove ABC  APC (cont.) • 1 + 3 = 2 + 4 • By the Betweenness of Rays Theorem, • 1 + 3 = ABC • 2 + 4 = APC • Substitution gives ABC = APC.

  15. Part II: Prove ABC  APC (cont.) • AB = AP, ABC = APC, and BC = PC… • …so ABC  APC by SAS.

  16. Part III: Conclusion • Because ABC  APC and APC  DFE, by transitivity of “”, ABC  DFE.

  17. Assignments • For Thursday/Friday:Asg #28 • For Monday, 11/25: Unit 3 Test Analysis • For Tuesday, 11/26: Asg #29

  18. Clean-up / Reminders • Pick up all trash / items. • Push in chairs (at front and back tables). • See you tomorrow!

  19. Proof, alternate version • The following slides contain a more condensed presentation of the SSS Theorem’s proof.

  20. Part II: Prove ABC  APC Given: 1. AB = DF 3. AC = DE 2. BC = FE Statement Reason Prove: ABC  APC. • PAC = EDF • AP = DF • APC  DFE • PC = EF • PC = BC • AP = AB • Protractor Postulate • Ruler Postulate. • SAS with 5, 4, 3. • CPCTE with 6. • Substitution, 2, 3.(or, transitivity of =) • Substitution, 1, 5.

  21. Part II: Prove ABC APC (cont.) Given: 1. AB = DF 3. AC = DE 2. BC = FE Statement Reason Prove: ABC  APC. • 2 = 1 • 3 = 4 • 2 + 4 = 1 + 3 • 2 + 4 = APC • 1 + 3 = ABC • APC = ABC • Theorem 9 (AP = AB) • Theorem 9 (PC = BC) • Addition (10 + 11) • Angle Addition Thm. • Angle Addition Thm. • Substitution on 12 using 13 and 14.

  22. Part II: Prove ABC APC (cont.) Given: 1. AB = DF 3. AC = DE 2. BC = FE Statement Reason Prove: ABC  APC. • APC  ABC • SAS with 8, 15, 9: 8: PC = BC 15: APC = ABC 9: AP = AB Part III : Conclusion Because ABC  APC and APC  DFE, by transitivity of “”, ABC  DFE.

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