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2.4 – Building a System of Geometry Knowledge

2.4 – Building a System of Geometry Knowledge. Euclid (325 BC – 265 BC). It’s all Greek to me!. Greek mathematician Lived ≈ 300 BC in Alexandria, Egypt Organized existing knowledge of geometry Wrote “The Elements” starting with:. 5 basic postulates. Addition Subtraction Division

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2.4 – Building a System of Geometry Knowledge

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  1. 2.4 – Building a System of Geometry Knowledge

  2. Euclid (325 BC – 265 BC)

  3. It’s all Greek to me! • Greek mathematician • Lived ≈ 300 BC in Alexandria, Egypt • Organized existing knowledge of geometry • Wrote “The Elements” starting with:

  4. 5 basic postulates • Addition • Subtraction • Division • Multiplication • Substitution

  5. Properties • Addition Property • You can add the same thing to both sides • If a = b then a + c = b + c

  6. Properties • Subtraction Property • You can subtract the same thing from both sides • If a = b then a – c = b – c

  7. Properties • Multiplication Property • You can multiply the same thing to both sides • If a = b then ac = bc

  8. Properties • Division Property • You can divide both sides by the same thing • If a = b then a / c = b / c (c cannot = 0 because you cannot ever divide by 0)

  9. Properties • Substitution Property • If a = b then you can replace a with b or bwith a in any equation

  10. Properties • Reflexive property (1 thing) • Anything is equal/congruent to itself • A = A • Symmetric Property (2 things) • If A = B, then B = A • Transitive Property (3 or more things) • If A = B and B = C, then A = C

  11. Properties • Distributive Property • a(b + c) = ab + ac • (b + c)a = ab + ac (just to see if you remember!)

  12. THEOREM a statement that has been proven deductively

  13. B C D A Theorems • Overlapping Segments Theorem • Given a segment with points A, B, C and D arranged consecutively and if AB = CD then AC = BD and if AC = BD then AB = CD

  14. B C A D O Theorems • Overlapping Angles Theorem • Given angle AOD with points B and C in the interior, then if mAOB = mCOD then mAOC = mBOD and if mAOC = mBOD then mAOB = mCOD

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