Table of Contents | Introduction | Home P l a n e G e o m e t r y An Adventure in Language and Logic based on CONSTRUCTION
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1. | From a given point to draw a straight line equal to a given straight line. | |||
2. | Let A be the given point, and BC the given straight line; | |||
3. | we are required to draw from the point A a straight line equal | |||
4. | to BC. | |||
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(The construction begins by drawing the equilateral triangle ABD; then extending DA, DB, to E and F; drawing a circle with radius BC, thus making BG equal to BC; then making DG, DL radii of a circle, so that on subtracting DB, DA respectively, AL will equal BG and therefore BC.) |
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5. | From the point A to the point B, draw the straight line AB; | |||
(Postulate 1) |
6. | and on it draw the equilateral triangle ABD; | (I. 1) | |||
7. | and extend the straight lines DA, DB to E, F. | (Postulate 2) | |||
8. | With B as center and BC as radius, draw the circle CHG, | ||||
9. | meeting DF at G. | (Postulate 3) | |||
10. | With D as center and DG as radius, draw the circle GKL, | ||||
11. | meeting DE at L. | (Postulate 3) | |||
12. | Then AL will equal BC. |
13. | For, since B is the center of the circle CHG, | |||
14. | BC is equal to BG. | (Definition 16) | ||
15. | And since D is the center of circle GKL, | |||
16. | DL is equal to DG. | |||
17. | But in those lines, DA is equal to DB; | (Definition 9) | ||
18. | therefore the remainder AL is equal to the remainder BG. | |||
(Axiom 3) | ||||
19. | And we have shown that BC is equal to BG; | |||
20. | therefore each of the straight lines AL, BC is equal to BG. | |||
21. | And things equal to the same thing are equal to one another. | |||
(Axiom 1) | ||||
22. | Therefore AL is equal to BC. | |||
23. | Therefore from the given point A we have drawn a straight line | |||
24. | AL equal to the straight line BC. | |||
25. | Which is what we wanted to do. |
This proposition leads directly to the next one, where we will be required to cut off from the longer of two straight lines a length equal to the shorter line. The solution is obvious -- but notice how we must rely on Proposition 2; line 6 below. In fact, the next proposition is the only one that requires Proposition 2. The "given" point is the endpoint of a line.
The student should now begin to see how each proposition depends on previous propositions. That is the nature of any logical theory. That is the axiomatic method.
1. | Given two unequal straight lines, to cut off from the longer line | |
2. | a straight line equal to the shorter line. | |
3. | Let AB and C be the two given straight lines, and let AB be | |
4. | longer; | |
5. | we are required to cut off from AB a straight line equal to C. | |
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6. | From the point A draw AD equal to C; | (Proposition 2) |
7. | and with A as center and radius AD, draw the circle DEF. | |
(Postulate 3) | ||
8. | Then, since the point A is the center of circle DEF, | |
9. | AE is equal to AD. | (Definition 16) |
10. | But C is also equal to AD. | (Construction) |
11. | Therefore each of the lines AE, C is equal to AD. | |
12. | Therefore AE is equal to C. | (Axiom 1) |
13. | Therefore from AB the longer of two straight lines | |
14. | we have cut off a straight line AE equal to C, the shorter line. | |
15. | Which is what we wanted to do. |
Please "turn" the page and do some Problems.
or
Continue on to the next proposition.
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