phi excursion 4

phi excursion 4 - AB and line BC is equal to line BA then...

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Rebecca Kuemmel October 20, 2008 PHI 108.04 Word Count: 226 Excursion 4: Euclid Part I. 1. AB is a given finite straight line. 2. Construct an equilateral triangle on AB . 3. Circle BCD can be described with centre A [Def. 16] and distance AB [Post. 3]. 4. Circle ACE can be described with centre B [Def. 16] and distance BA [Post. 3]. 5. From point C (where the two circles cross each other) to the points A and B , let the straight lines CA and CB be joined. [Def. 3, Post. 1] 6. Since point A is the centre of circle BCD , straight lines AC and AB are equal [Def. 15] 7. Since point B is the centre of circle ACE , straight lines BC and BA are equal [Def. 15] 8. If line AC is equal to line
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Unformatted text preview: AB and line BC is equal to line BA, then lines AC and BC are equal. [C.N. 1] 9. Therefore lines AB, BC and CA are equal to one another. [C.N. 1] 10. Therefore triangle ABC is and equilateral triangle constructed on straight line AB. [Def. 20] Part II. The proof makes the assumption that the two circles cross at exactly a point where they pass directly over the centre of the other circle which would make this proof correct. Also, the proof assumes that the two circles are the same size. If the circles were not the same size or they did not pass directly over the other’s centre, this proof would be incorrect....
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